Rd Sharma Xi 2018 Solutions for Class 12 Science Math Chapter 12 Mathematical Induction are provided here with simple step-by-step explanations. These solutions for Mathematical Induction are extremely popular among Class 12 Science students for Math Mathematical Induction Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 12 Science Math Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 12.27:

Question 1:

1 + 2 + 3 + ... + nn(n+1)2 i.e. the sum of the first n natural numbers is n(n+1)2.

Answer:

Let P(n) be the given statement.
Now,
P(n) = 1 + 2 + 3 +...+ n  =n(n+1)2
Step 1:     P(1) =1 =1(1+1)2=1Hence, P(1) is true.Step 2: Let P(m) be true.Then,1+2+3+...+m=m(m+1)2We shall now prove that P(m+1) is true.i.e., 1+2+...+(m+1) =(m+1)(m+2)2Now, 1+2+...+m =m(m+1)21+2+...m+m+1 =m(m+1)2+m+1          Adding m+1 to both sides                                    =m2+m+2m+22                                    =(m+1)(m+2)2Hence, P(m+1) is true.                         
By the principle of mathematical induction, P(n) is true for all nN.

Page No 12.27:

Question 2:

12 + 22 + 32 + ... + n2 = n(n + 1)(2n + 1)6

Answer:

Let P(n) be the given statement.
Now,

P(n) =12+22+32+...+n2=n(n+1)(2n+1)6Step 1:P(1) =12=1(1+1)(2+1)6=66=1Hence, P(1) is true.Step 2:Let P(m) be true. Then,12+22+...+m2=m(m+1)(2m+1)6We shall now prove that P(m+1) is true.i.e.,  12+22+32+...+(m+1)2=(m+1)(m+2)(2m+3)6Now,P(m) = 12+22+32+...+m2=m(m+1)(2m+1)612+22+32+...+m2+(m+1)2=m(m+1)(2m+1)6+(m+1)2          Adding (m+1)2 to both sides12+22+32+...+(m+1)2=m(m+1)(2m+1)+6(m+1)26=(m+1)(2m2+m+6m+6)6=(m+1)(m+2)(2m+3)6Hence, P(m+1) is true.By the principle of mathematical induction, the given statement is true for all nN.

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Question 3:

1 + 3 + 32 + ... + 3n−1 = 3n-12

Answer:

Let P(n) be the given statement.
Now,
P(n) =1+3+32+...+3n-1=3n-12Step 1:P(1) =1 =31-12=22=1Hence, P(1) is true.Step 2:Let P(m) is true.Then,1+3+32+...+3m-1=3m-12We shall prove that P(m+1) is true.That is,   1+3+32+...+3m=3m+1-12Now, we have:1+3+32+...+3m-1=3m-121+3+32+...+3m-1+3m=3m-12+3m            Adding 3m to both sides1+3+32+...+3m=3m-1+2×3m2=3m(1+2)-12=3m+1-12Hence, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 4:

11.2+12.3+13.4+...+1n(n+1)=nn+1

Answer:

Let P(n) be the given statement.
Now,
P(n) =11.2+12.3+13.4+...+1n(n+1)=nn+1Step 1:P(1) =11.2=12=11+1Hence, P(1) is true.Step 2:Let P(m) be true.Then,11.2+12.3+13.4+...+1m(m+1)=mm+1We shall now prove that P(m+1) is true.i.e.,   11.2+12.3+13.4+...+1(m+1)(m+2)=m+1m+2Now,P(m)= 11.2+12.3+13.4+...+1m(m+1)=mm+111.2+12.3+13.4+...+1m(m+1)+1(m+1)(m+2)=mm+1+1(m+1)(m+2)                Adding 1(m+1)(m+2)  to both sides11.2+12.3+13.4+...+1(m+1)(m+2)=m2+2m+1(m+1)(m+2)=(m+1)2(m+1)(m+2)=m+1m+2Therefore, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 5:

1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.

Answer:

Let P(n) be the given statement.
Now,
P(n) = 1+3+5+...+(2n-1)=n2Step 1:P(1) =1 = 12Hence, P(1) is true.Step 2 :Let P(m) be true.Then,1+3+5+...+(2m-1) = m2To prove: P(m+1) is true.i.e.,  1+3 +5+...+2m+1-1=m+12 1+3 +5+...+2m+1=m+12Now, we have:1+3+5+...+(2m-1) =m21+3+...+(2m-1)+(2m+1) =m2+2m+1             Adding 2m+1 to both sides1+3+5+...+(2m+1) =(m+1)2Hence, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 6:

12.5+15.8+18.11+...+1(3n-1)(3n+2)=n6n+4

Answer:

Let P(n) be the given statement.
Now,
P(n) =12.5+15.8+18.11+...+1(3n-1)(3n+2)=n6n+4Step 1:P(1) =12.5=110=16+4Hence, P(1) is true.Step 2:Let P(m) be true. Then,12.5+15.8+18.11+...+1(3m-1)(3m+2)=m6m+4To prove: P(m+1) is true.i.e.,12.5+15.8+...+1(3m+2)(3m+5)=m+16m+10Thus, we have: 12.5+15.8+18.11+...+1(3m-1)(3m+2)=m6m+412.5+15.8+...+1(3m-1)(3m+2)+1(3m+2)(3m+5)=m6m+4+1(3m+2)(3m+5)            Adding 1(3m+2)(3m+5) to both sides12.5+15.8+...+1(3m+2)(3m+5)=3m2+5m+22(3m+2)(3m+5)=(3m+2)(m+1)2(3m+2)(3m+5)=m+16m+10Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 7:

11.4+14.7+17.10+...+1(3n-2)(3n+1)=n3n+1

Answer:

Let P(n) be the given statement.
Now,
P(n) =11.4+14.7+17.10+...+1(3n-2)(3n+1)=n3n+1Step 1:P(1) =11.4=14=13×1+1Hence, P(1) is true.Step 2:Let P(m) be true.i.e.,11.4+14.7+...+1(3m-2)(3m+1)=m3m+1To prove: P(m+1) is true.i.e.,11.4+14.7+...+1(3m+1)(3m+4)=m+13m+4Now,P(m)=11.4+14.7+...+1(3m-2)(3m+1)=m3m+111.4+14.7+...+1(3m-2)(3m+1)+1(3m+1)(3m+4)=m3m+1+1(3m+1)(3m+4)              Adding 1(3m+1)(3m+4) to both sides11.4+14.7+...+1(3m+1)(3m+4)=3m2+4m+1(3m+1)(3m+4)=(3m+1)(m+1)(3m+1)(3m+4)=m+13m+4Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

Page No 12.27:

Question 8:

13.5+15.7+17.9+...+1(2n+1)(2n+3)=n3(2n+3)

Answer:

Let P(n) be the given statement.
Now,
P(n)=13.5+15.7+17.9+...+1(2n+1)(2n+3)=n3(2n+3)Step 1: P(1)=13.5=115=13(2+3)Hence, P(1) is true.Step 2:Let P(m) be true. Then,13.5+15.7+17.9+...+1(2m+1)(2m+3)=m3(2m+3)To prove: P(m+1) is true.That is,13.5+15.7+17.9+...+1(2m+3)(2m+5)=m+13(2m+5)Now, P(m) = 13.5+15.7+17.9+...+1(2m+1)(2m+3)=m3(2m+3)13.5+15.7+...+1(2m+1)(2m+3)+1(2m+3)(2m+5)=m3(2m+3)+1(2m+3)(2m+5)           Adding 1(2m+3)(2m+5) to both sides13.5+15.7+...+1(2m+3)(2m+5)=2m2+5m+33(2m+3)(2m+5)=(2m+3)(m+1)3(2m+3)(2m+5)=m+132m+5Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 9:

13.7+17.11+111.5+...+1(4n-1)(4n+3)=n3(4n+3)

Answer:

Let P(n) be the given statement.
Now,
P(n) =13.7+17.11+111.15+...+1(4n-1)(4n+3)=n3(4n+3)Step 1:P(1) =13.7=121=13(4+3)Hence, P(1) is true.Step 2:Let P(m) is true. Then,13.7+17.11+...+1(4m-1)(4m+3)=m3(4m+3)To prove: P(m+1) is true.That is,13.7+17.11+...+1(4m+3)(4m+7)=m+13(4m+7)Now,P(m) =13.7+17.11+...+1(4m-1)(4m+3)=m3(4m+3)13.7+17.11+...+1(4m-1)(4m+3)+1(4m+3)(4m+7)=m3(4m+3)+1(4m+3)(4m+7)                Adding 1(4m+3)(4m+7) to both sides13.7+17.11+...+1(4m+3)(4m+7)=4m2+7m+33(4m+3)(4m+7)=(4m+3)(m+1)3(4m+3)(4m+7)=m+13(4m+7)Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 10:

1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2

Answer:

Let P(n) be the given statement.
Now,
P(n) =1.2+2.22+3.23+...+n.2n=(n-1)2n+1+2Step 1:P(1)=1.2=2=(1-1)21+1+2Thus, P(1) is true.Step 2:Let P(m) be true.Then,1.2+2.22+...+m.2m=(m-1)2m+1+2To prove: P(m+1) is true.That is,1.2+2.22+...+(m+1)2m+1=m.2m+2+2Now, P(m) =1.2+2.22+...+m.2m=(m-1)2m+1+21.2+2.22+...+m.2m+(m+1).2m+1=(m-1)2m+1+2+(m+1).2m+1             Adding (m+1).2m+1 to both sidesP(m+1)=2m.2m+1+2=m.2m+2+2Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

Page No 12.27:

Question 11:

2 + 5 + 8 + 11 + ... + (3n − 1) =12n(3n+1)

Answer:

Let P(n) be the given statement.
Now,
P(n) =2+5+8+...+(3n-1)=12n(3n+1)Step 1:P(1)=2=12×1(3+1) Hence, P(1) is true.Step 2:Let P(m) be true.Then,2+5+8+...+(3m-1)=12m(3m+1)To prove: P(m+1) is true.That is,2+5+8+...+(3m+2)=12(m+1)(3m+4)P(m) is equal to:2+5+8+...+(3m-1)=12m(3m+1)Thus, we have:2+5+8+...+(3m-1)+(3m+2)=12m(3m+1)+(3m+2)                Adding (3m+2) to both sides2+5+8+...+(3m+2)=12(3m2+m+6m+4)=12(3m2+7m+4)2+5+8+...+(3m+2)=12(3m+4)(m+1)Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 12:

1.3 + 2.4 + 3.5 + ... + n. (n + 2) =16n(n+1)(2n+7)

Answer:

Let P(n) be the given statement.
Now,
P(n) = 1.3+2.4+3.5+...+n.(n+2)=16n(n+1)(2n+7)Step 1:P(1)= 1.3 =3 = 16×1(1+1)(2×1+7)Hence, P(1) is true.Step 2:Let P(m) be true.Then,1.3+2.4+...+m.(m+2)=16m(m+1)(2m+7)To prove: P(m+1) is true.That is,1.3+2.4+...+(m+1)(m+3)=16(m+1)(m+2)(2m+9)P(m) is equal to 1.3+2.4+...+m(m+2)=16m(m+1)(2m+7).Thus, we have:1.3+2.4+...+m(m+2)+(m+1)(m+3) = 16m(m+1)(2m+7)+(m+1)(m+3)            Adding (m+1)(m+3) to both sides1.3+2.4+...+(m+1)(m+3)=16(m+1)2m2+7m+6m+18                                                             =16(m+1)(2m2+13m+18)                                                             =16(m+1)(2m+9)(m+2)Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 13:

1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) = n(4n2+6n-1)3

Answer:

Let P(n) be the given statement.
Now,
P(n)=1.3+3.5+5.7+...+(2n-1)(2n+1)=n(4n2+6n-1)3Step 1:P(1)=1.3 =3 =1(4×12+6×1-1)3Hence, P(1) is true.Step 2:Let P(m) be true.Then,1.3+3.5+...+(2m-1)(2m+1)=m(4m2+6m-1)3To prove: P(m+1) is true.That is,1.3+3.5+...+(2m+1)(2m+3)=(m+1)4(m+1)2+6m+1-13Now, P(m) is equal to: 1.3+3.5+...+(2m-1)(2m+1)=m(4m2+6m-1)31.3+3.5+...+(2m-1)(2m+1)+(2m+1)(2m+3)=m(4m2+6m-1)3+(2m+1)(2m+3)            Adding (2m+1)(2m+3) to both sidesP(m+1)=m(4m2+6m-1)+3(4m2+8m+3)3P(m+1)=4m3+6m2-m+12m2+24m+93=4m3+18m2+23m+93P(m+1)=4m(m2+2m+1)+10m2+19m+93                   =4m(m+1)2+(10m+9)(m+1)3                   =(m+1)4m(m+1)+10m+93                   =(m+1)3(4m2+8m+4+6m+5)                   =(m+1)4(m+1)2+6m+1-13Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 14:

1.2 + 2.3 + 3.4 + ... + n (n + 1) = n(n+1)(n+2)3

Answer:

Let P(n) be the given statement.
Now,
P(n)=1.2+2.3+3.4+...+n(n+1)=n(n+1)(n+2)3Step 1:P(1)=1.2=2=1(1+1)(1+2)3Hence, P(1) is true.Step 2:Let P(m) be true.Then,1.2+2.3+...+m(m+1)=m(m+1)(m+2)3To prove: P(m+1) is true.That is,1.2+2.3+...+(m+1)(m+2)=(m+1)(m+2)(m+3)3Now, P(m) is 1.2+2.3+...+m(m+1)=m(m+1)(m+2)31.2+2.3+...+m(m+1)+(m+1)(m+2)=m(m+1)(m+2)3+(m+1)(m+2)P(m+1)=(m+1)(m+2)(m+3)3Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.'

Page No 12.27:

Question 15:

12+14+18+...+12n=1-12n

Answer:

Let P(n) be the given statement.
Now,
P(n): 12+14+18+...+12n=1-12nStep 1: P(1)=12=1-121Thus, P(1) is true.Step 2:Suppose P(m) is true.Then,12+14+...+12m=1-12mTo show: P(m+1) is true whenever P(m) is true.That is,12+14+...+12m+1=1-12m+1Now, P(m) is true.Thus, we have:12+14+...+12m=1-12m12+14+...+12m+12m+1=1-12m+12m+1          Adding 12m+1 to both sidesP(m+1)=1-12m+12m.2=1-12 m1-12=1-12m+1Thus, P(m+1) is true.
By the principle of mathematical induction, P(n) is true for all nN.

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Question 16:

12 + 32 + 52 + ... + (2n − 1)2 = 13n(4n2-1)

Answer:

Let P(n) be the given statement.
Now,
P(n)=12+32+52+...+(2n-1)2=13n(4n2-1)Step 1:  P(1)=12=1=13×1×(4-1)Hence, P(1) is true.Step 2:Let P(m) be true.Then,12+32+...+(2m-1)2=13m(4m2-1)To prove: P(m+1) is true whenever P(m) is true.That is, 12+32=...+(2m+1)2=13(m+1)4(m+1)2-1We know that P(m) is true.Thus, we have:12+32+...+(2m-1)2=13m(4m2-1)12+32+...+(2m-1)2+(2m+1)2=13m(4m2-1)+(2m+1)2                Adding (2m+1)2 to both sidesP(m+1)=134m3-m+12m2+12m+3P(m+1)=13(4m3-m+8m2+4m+4m2+8m+3)                     =13(m+1)(4m2+8m+3)                    =13(m+1)(4(m+1)2-1)Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 17:

a + ar + ar2 + ... + arn−1 = arn-1r-1,r1

Answer:

Let P(n) be the given statement.
Now,
P(n)= a+ar+ar2+...+arn-1=arn-1r-1, r1Step 1:P(1)=a=ar1-1r-1Hence, P(1) is true.Step 2:Suppose P(m) is true. Then,a+ar+ar2+...+arm-1=arm-1r-1, r1To show: P(m+1) is true whenever P(m) is true.That is,a+ar+ar2+...+arm=arm+1-1r-1, r1We know that P(m) is true.Thus, we have:a+ar+ar2+...+arm-1=arm-1r-1a+ar+ar2+...+arm-1+arm=arm-1r-1+arm          Adding arm to both sidesP(m+1)=arm-1+r.rm-rmr-1P(m+1)=arm+1-1r-1,  r1Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.



Page No 12.28:

Question 18:

a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = n22a+(n-1)d

Answer:

Let P(n) be the given statement.
Now,
P(n): a+(a+d)+(a+2d)+...+(a+(n-1)d)=n22a+(n-1)dStep1:P(1)= a= 12(2a+(1-1)d)Hence, P(1) is true.Step 2:Suppose P(m) is true.Then,a+(a+d)+...+(a+(m-1)d)=m22a+(m-1)dWe have to show that P(m+1) is true whenever P(m) is true.That is,a+(a+d)+...+(a+md)=(m+1)22a+mdWe know that P(m) is true.Thus, we have:a+(a+d)+...+(a+(m-1)d)=m22a+(m-1)da+(a+d)+...+(a+(m-1)d)+(a+md)=m22a+(m-1)d+(a+md)          Adding (a+md) to both sidesP(m+1)=122am+m2d-md+2a+2mdP(m+1)=122a(m+1)+md(m+1)                   =12(m+1)(2a+md)Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 19:

52n −1 is divisible by 24 for all nN.

Answer:

Let P(n) be the given statement.
Now,
P(n): 52n-1 is divisible by 24 for all nN.Step 1: P(1)=52-1=25-1=24 It is divisible by 24.Thus, P(1) is true.Step 2:Let P(m) be true.Then, 52m-1 is divisible by 24.Now, let 52m-1 = 24λ, where λN.We need to show that P(m+1) is true whenever P(m) is true.Now,P(m+1) =52m+2-1               =52m52-1               =25(24λ+1) -1               =600λ+24               =24(25λ+1)It is divisible by 24.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 20:

32n+7 is divisible by 8 for all nN.

Answer:

Let P(n) be the given statement.
Now,
P(n): 32n+7 is divisible by 8 for all nN.Step 1:P(1)= 32+7=9+7=16 It is divisible by 8.Step 2:  Let P(m) be true.Then, 32m+7 is divisible by 8.Thus, 32m+7=8λ for some λN.                ...(1)We need to show that P(m+1) is true whenever P(m) is true.Now,P(m+1) =32m+2+7                =32m.9 +7                =(8λ-7).9+7             From (1)                 =72λ-63+7                 =72λ-56                 =8(9λ-7)   It is a multiple of 8.Thus, P(m+1) is divisible by 8.By the principle of mathematical induction, P(n) is true for all nN.

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Question 21:

52n+2 −24n −25 is divisible by 576 for all nN.

Answer:

Let P(n) be the given statement.
Now,
P(n): 52n+2-24n-25 is divisible by 576 for all nN.Step 1:P(1)= 52+2-24-25=625-49=576 It is divisible by 576.Thus, P(1) is true.Step2: Let P(m) be true.Then,52m+2-24m-25 is divisible by 576.Let  52m+2-24m-25 =576λ, where λN.We need to show that P(m+1) is true whenever P(m) is true.Now, P(m+1)=52m+4-24(m+1)-25=52×(576λ+24m+25)-24m-49=25×576λ+600m+625-24m-49=25×576λ+576m+576=576(25λ+m+1) It is divisible by 576.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 22:

32n+2 −8n − 9 is divisible by 8 for all nN.

Answer:

Let P(n) be the given statement.
Now,
P(n): 32n+2-8n-9 is divisible by 8 for all nN.Step 1:P(1)=32+2-8-9=81-17=64 It is divisible by 8.Thus, P(1) is true.Step(2):Let P(m) be true.Then, 32m+2-8m-9 is divisible by 8.Let:32m+2-8m-9 =8λ where λN           ...(1)We need to show that P(m+1) is true whenever P(m) is true.Now,P(m+1)=32m+4-8m+1-17=(8λ+8m+9) -8m -8-17         From (1)=8λ -16 =8(λ-1) It is divisible by 8.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 23:

(ab)n = anbn for all nN.

Answer:

Let P(n) be the given statement.
Now,
P(n): (ab)n=anbn for all nN.Step 1:P(1):(ab)1=a1b1=abThus, P(1) is true.Step 2:Let P(m) be true.Then,(ab)m=ambmWe need to show that P(m+1) is true whenever P(m) is true.Now, P(m+1): (ab)m+1=(ab)m. ab                                        =ambm.ab                                        =ama.bmb                                        =am+1bm+1Hence, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 24:

n(n + 1) (n + 5) is a multiple of 3 for all nN.

Answer:

Let P(n) be the given statement.
Now,
P(n):n(n+1)(n+5) is a multiple of 3.Step1:P(1): 1(1+1)(1+5)=12 It is a multiple of 3.Hence, P(1) is true.Step2: Let Pm be true.Then, mm+1m+5 is a multiple of 3.Suppose mm+1m+5=3λ, where λN.We have to show that Pm+1 is true whenever P(m) is true.Now,P(m+1)=m+1m+2m+6              =mm+1m+6+2m+1m+6             =mm+1m+5+1+2m+1m+6             =mm+1m+5+m(m+1)+2m+1m+6             =3λ+m+1m+2m+6                                          From P(m)             =3λ+3m+1m+2It is clearly a multiple of 3.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 25:

72n + 23n−3. 3n−1 is divisible by 25 for all nN.

Answer:

Let P(n) be the given statement.
Now,
P(n): 72n+23n-3.3n-1  is divisible by 25.Step1:  P(1): 72+23-3.31-1=49+1=50 It is divisible by 25.Thus, P(1) is trueStep2:  Let Pm be true.Now,72m+23m-3.3m-1 is divisible by 25.Suppose: 72m+23m-3.3m-1= 25λ             ...(1)We have to show that Pm+1 is true whenever P(m) is true.Now, Pm+1=72m+2+23m.3m                        =72m+2+72.23m-3.3m-1-72.23m-3.3m-1+23m.3m                        =7272m+23m-3.3m-1+23m.3m1-4924                        =72×25λ-23m.3m×2523.31                 Using (1)                        =2549λ-23m-3.3m-1 It is divisible by 25.Thus, Pm+1 is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 26:

2.7n + 3.5n − 5 is divisible by 24 for all nN.

Answer:

Let P(n) be the given statement.
Now,
P(n): 2.7n+3.5n-5 is divisible by 24.Step1:P(1): 2.71+3.51-5 = 24 It is divisible by 24.Thus, P1 is true.Step2:  Let Pm be true.Then,2.7m+3.5m-5 is divisible by 24.Suppose:2.7m+3.5m-5=24λ             ...1We need to show that Pm+1 is true whenever Pm is true.Now, Pm+1=2.7m+1+3.5m+1-5                        =2.7m+1+(24λ+5-2.7m)5-5                        =2.7m+1+120λ+25-10.7m-5                        =2.7m.7-10.7m+120λ+24-4                      =7m 14-10+120λ+24-4                      =7m.4+120λ+24-4                      =47m-1+245λ+1                        =4×6μ+24(5λ+1)          Since 7m-1  is a multiple of 6 for all nN, 7m-1=μ.                        =24(μ+5λ+1)  It is a multiple of 24.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for nN.

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Question 27:

11n+2 + 122n+1 is divisible by 133 for all nN.

Answer:

Let P(n) be the given statement.
Now,
Pn: 11n+2+122n+1 is divisible by 133.Step1: P1=111+2+122+1=1331+1728=3059 It is divisible by 133.Step2:Let Pm be divisible by 133.Now,11m+2+122m+1 is divisible by 133.Suppose:11m+2+122m+1=133λ                    ...(1)We shall show that Pm+1 is true whenever Pm is true.Now, Pm+1=11m+3+122m+3                      =11m+2.11+122m+1.122+11.122m+1-11.122m+1                      =1111m+2+122m+1+122m+1144-11                      =11.133λ+122m+1.133                    From (1)                      =13311λ+122m+1  It is divisible by 133.Thus, Pm+1 is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 28:

Given a1=12a0+Aa0,a2=12a1+Aa1 and an+1=12an+Aan for n ≥ 2, where a > 0, A > 0.
Prove that an-Aan+A=a1-Aa1+A2n-1

Answer:

Let:Pn:an-Aan+A=a1-Aa1+A2n-1Step I  P(1):a1-Aa1+A=a1-Aa1+A21-1         (which is true)P(2):a2-Aa2+A=12a1+Aa1-A12a1+Aa1+A=a1+Aa1-2Aa1+Aa1+2A=a1+A-2Aa1+A+2A=a1-Aa1+A22-1Thus, P(1) and P(2)  are true.Step II  Let P(k) be true.Now, ak-Aak+A=a1-Aa1+A2k-1     .....(i)andP(k+1):ak+1-Aak+1+A=12ak+Aak-A12ak+Aak+A=ak+Aak-2Aak+Aak+2A=ak-Aak2ak+Aak2=ak-Aakak+Aak2=ak-Aak+A2=a1-Aa1+A2k-12=a1-Aa1+A2k=a1-Aa1+A2k+1-1 Thus, P(k+1) is also true.

=akak+Aak-Aak+A2k-1-Aak-Aak+A2=akak+Aak-Aak+A2k-1-Aak+Aak-Aak+A2k-1ak+A2    Using (i)=ak+Aak-Aak+A2k-1ak-Aak+A2=ak-Aak+A2k-1ak-Aak+A=ak-Aak+A2k-1ak-Aak+A=ak-Aak+A2k-1+1=a1-Aa1+A2k-12k-1+1=

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Question 29:

Prove that n3 - 7+ 3 is divisible by 3 for all n  N.

Answer:

Let pn=n3-7n+3 is divisible by 3 nN.Step I: For n=1,p1=13-7×1+3=1-7+3=-3, which is clearly divisible by 3So, it is true for n=1Step II: For n=k,Let pk=k3-7k+3=3m, where m is any integer, be true kN.Step III: For n=k+1,pk+1=k+13-7k+1+3=k3+3k2+3k+1-7k-7+3=k3+3k2-4k-3=k3-7k+3+3k2+3k-6=3m+3k2+k+2       Using step II=3m+k2+k+2=3p, where p is any integerSo, pk+1 is divisible by 3.

Hence, n3 - 7+ 3 is divisible by 3 for all n  N.

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Question 30:

Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n N.

Answer:

Let pn:1+2+22+...+2n=2n+1-1 nNStep I: For n=1,LHS=1+21=3RHS=21+1-1=22-1=4-1=3As, LHS=RHSSo, it is true for n=1.Step II: For n=k,Let pk:1+2+22+...+2k=2k+1-1 be true kNStep III: For n=k+1,LHS=1+2+22+...+2k+2k+1=2k+1-1+2k+1       Using step II=2×2k+1-1=2k+1+1-1=2k+2-1RHS=2k+1+1-1=2k+2-1As, LHS=RHSSo, it is also true for n=k+1.

Hence, 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n  N.

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Question 31:

7 + 77 + 777 + ... + 777  ...........n-digits7=781(10n+1-9n-10)

Answer:

Let P(n) be the given statement.
Now,
P(n): 7+77+777+...+777...n digits...7=781(10n+1-9n-10)Step(1): P(1)  = 7 =781(102-9-10)=781×81 Thus, P(1) is true.Step 2:  Let P(m) be true.Then, 7+77+777+...+777...m digits...7=781(10m+1-9m-10)We need to show that P(m+1) is true whenever P(m) is true.

Now, P(m + 1) = 7 + 77 + 777 +....+ 777...(m + 1) digits...7

This is a geometric progression with n= m+1.Sum P(m+1): =799+99+999+...m+1term=7910-1+100-1+...(m+1) term=7910+100+1000+...(m+1) term  -(1+1+1...m+1 times...+1=791010m+1-19-m+1=78110m+2-9m-19Thus, P(m+1) is true.By the principle of mathematical induction, Pn is true for all nN.

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Question 32:

n77+n55+n33+n22-37210n is a positive integer for all n ∈ N.

Answer:

Let P(n) be the given statement.
Now,
P(n): n77+n55+n33+n22-37210n is a positive integer.Step 1:    P(1)=17+15+13+12-37210=30+42+70+105-37210=210210=1 It is a positive integer.Thus, P(1) is true.Step 2:Let P(m) be true.Then, m77+m55+m33+m22-37210m is a positive integer.Let m77+m55+m33+m22-37210m=λ for some λpositive N.To show: Pm+1 is a positive integer.Now,P(m+1)=m+177+m+155+m+133+m+122-37210m+1=17m7+7m6+21m5+35m4+35m3+21m2+7m+1+15m5+5m4+10m3+10m2+5m+1+13m3+3m2+3m+1+12m2+2m+1-37210m-37210           =m77+m55+m33+m22-37210m +m6+3m5+6m4+7m3+6m2+4m=λ+  m6+3m5+6m4+7m3+6m2+4mIt is a positive integer, as λ is a positive integer.Thus, Pm+1 is true,By the principle of mathematical induction, P(n) is true for all nN.                             

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Question 33:

n1111+n55+n33+62165n is a positive integer for all nN.

Answer:

Let P(n) be the given statement.
Now,
P(n): n1111+n55+n33+62165n is a positive integer for all nN.Step 1:P(1) =111+15+13+62165=15+33+55+62165=165165=1 It is certainly a positive integer.Hence, P(1) is true.Step2:Let P(m) be true.Then, m1111+m55+m33+62165m is a positive integer.Now, let m1111+m55+m33+62165m =λ,  where λN is a positive integer.We have to show that P(m+1) is true whenever P(m) is true.To prove: (m+1)1111+(m+1)55+(m+1)33+62165(m+1) is a positive integer.Now,(m+1)1111+(m+1)55+(m+1)33+62165(m+1)=111m11+11m10+55m9+165m8+330m7+462m6+462m5+330m4+165m3+55m2+11m+1+15m5+5m4+10m3+10m2+5m+1+13m3+3m2+3m+1+62165m+62165=m1111+m55+m33+62165m+m10+5m9+15m8+30m7+42m6+42m5+31m4+17m3+8m2+3m+111+15+13+6105=λ+m10+5m9+15m8+30m7+42m6+42m5+31m4+17m3+8m2+3m+1It is a positive integer.Thus, P(m+1) is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 34:

12tanx2+14tanx4+...+12ntanx2n=12ncotx2n-cot x for all nN and 0<x<π2

Answer:

We need to prove 12tanx2+14tanx4+...+12ntanx2n=12ncotx2n-cot x for all nN and 0<x<π2 using mathematical induction.

For n = 1,

LHS = 12tanx2

and

 RHS=12cotx2-cotx=12tanx2-1tanxRHS=12tanx2-12tanx21-tan2x2RHS=12tanx2-1-tan2x22 tanx2=1-1+tan2x22tanx2=tanx22

Therefore, the given relation is true for n = 1.

Now, let the given relation be true for n = k.

We need to prove that the given relation is true for n = k + 1.

12tanx2+14tanx4+...+12ktanx2k=12kcotx2k-cot x

Now,

12tanx2+14tanx4+...+12ktanx2k+12k+1tanx2k+1=12kcotx2k-cot x+12k+1tanx2k+1

Let:

L=12kcotx2k-cot x+12k+1tanx2k+1.

L=12kcotx2k+12k+1tanx2k+1-cot xL=12ktanx2k+12k+1tanx2k+1-cot xL=12ktan2x2k+1+12k+1tanx2k+1-cot xL=12k×2tanx2k+11-tan2x2k+1+12k+1tanx2k+1-cot xL=1-tan2x2k+12k+1tanx2k+1+12k+1tanx2k+1-cot x

L=1-tan2x2k+1+tan2x2k+12k+1tanx2k+1-cot x=12k+1cotx2k+1-cot x
Now,

12tanx2+14tanx4+...+12ktanx2k+12k+1tanx2k+1=12k+1cotx2k+1-cot x

Thus, 12tanx2+14tanx4+...+12ntanx2n=12ncotx2n-cot x for all nN and 0<x<π2.

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Question 35:

Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all nN?

Answer:

Pn:2n3nWe know that Pr is true.Thus, we have:2r3rTo show: P(r+1) is true.We know:P(r) is true.2r3r2r.23r.2           Multiplying both sides by 22r+16r2r+13r+3r=2r+13r+3        Since 3r3 for all rN=2r+13r+1   Hence, P(r+1) is true.However, we cannot conclude that Pn is true for all nN.P(1): 213.1Therefore, Pn is not true for all nN.

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Question 36:

(2n)!22n(n!)213n+1 for all nN

Answer:

Let P(n) be the given statement.
Thus, we have:

Pn:2n!22nn!213n+1Step1:  P(1): 2!22.1=1213+1Thus, P(1) is true.Step2:   Let P(m) be true. Thus, we have:2m!22mm!213m+1We need to prove that P(m+1) is true. 

Now,

P(m+1): (2m+2)!22m+2(m+1)!2=2m+22m+12m!22m.22m+12m!2(2m+2)!22m+2(m+1)!22m!22mm!2×2m+22m+122m+12(2m+2)!22m+2(m+1)!22m+12m+13m+1

2m+2!22m+2m+1!22m+124m+123m+12m+2!22m+2m+1!24m2+4m+1×3m+443m3+7m2+5m+13m+42m+2!22m+2m+1!212m3+28m2+19m+412m3+28m2+20m+43m+412m3+28m2+19m+412m3+28m2+20m+4<12m+2!22m+2m+1!2<13m+4

Thus, P(m + 1) is true.

Hence, by mathematical induction (2n)!22n(n!)213n+1 is true for all nN

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Question 37:

1+14+19+116+...+1n2<2-1nfor all n ≥ 2, nN

Answer:

Let P(n) be the given statement.
Thus, we have:
Pn:1+14+19+...+1n2<2-1nStep1: P(2):122=14<2-12Thus, P2 is true.          We have not taken n =1 because it is not possible. We will start this function from n=2 onwards.Step2:  Let Pm be true. Now,1+14+19+...+1m2<2-1mWe need to prove that P(m+1) is true.We know that P(m) is true.Thus, we have:1+14+19+...+1m2<2-1m1+14+19+...+1m2+1m+12<2-1m+1m+12           Adding 1(m+1)2 to both sidesPm+1<2-1m+1       m+12>m+1, 1m+12<1m+11m-1m+12<1m+1 as m<m+1                   Thus, Pm+1  is true.By principle of mathematical induction, P(n) is true for all nN, n2.

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Question 38:

x2n−1 + y2n−1 is divisible by x + y for all nN.

Answer:

Let P(n) be the given statement.
Now,
P(n): x2n-1+y2n-1 is divisible by x+y.Step1:P(1):x2-1+y2-1=x+y is divisible by x+yStep2:Let P(m) be true.Also,x2m-1+y2m-1 is divisible by x+y.Suppose: x2m-1+y2m-1=λx+y where λN            ...(1)We shall show that Pm+1 is true whenever Pm is true.Now, Pm+1=x2m+1+y2m+1                        =x2m+1+y2m+1-x2m-1.y2+x2m-1.y2                        =x2m-1x2-y2+y2x2m-1+y2m-1                From (1)                        =x2m-1x2-y2+y2.λx+y                                  =x+yx2m-1x-y+λy2                 [It is divisible by (x+y).]Thus, Pm+1 is true.By the principle of mathematical induction, P(n) is true for all nN.

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Question 39:

sin x+ sin 3x+...+sin (2n-1)x=sin2nxsin x

Answer:

Let P(n) be the given statement.
P(n): sin x+sin 3x+...+sin2n-1x=sin2nxsin xStep 1:P(1): sinx=sin2xsinx Thus, P(1) is true. Step 2:Let P(m) be true. sin x+sin 3x+...+sin2m-1x=sin2mxsin xWe shall show that P(m+1) is true. We know that P(m) is true.  sin x+sin 3x+...+sin (2m-1) =sin2mxsin xsin x+sin 3x+...sin (2m-1)x+ sin (2m+1)x=sin2mxsin x+sin (2m+1)x      Adding sin (2m+1)x to both the sidesP(m+1)x=sin2mx+sin xsin mxcosm+1x+sinm+1x cosmxsin x                      =sin2mx+sin xsin mxcos mxcos x-sin2mxsinx+sin mxcos xcos mx+cos2mxsin xsin x                     =sin2mx+2sin xcos xcos mx-sin2x sin2mx+cos2mx sin2xsin x                     =sin2mx1-sin2x+2sin xcos xcos mx+cos2mx sin2xsin x                     =sin2mxcos2x+2sin xcos xcos mx+cos2mx sin2xsin x                     =sin mx cos x+cos mx sinx2sin x                     =sinm+12sin xHence, P(m+1) is true.  By the principle of mathematical induction, the given statement P(n) is true for all nN.



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Question 40:

Prove that cosα+cosα+β+cosα+2β+...+cosα+n-1β=cosα+n-12βsinnβ2sinβ2 for all nN.                     [NCERT EXEMPLAR]

Answer:

Let pn: cosα+cosα+β+cosα+2β+...+cosα+n-1β=cosα+n-12βsinnβ2sinβ2  nN.Step I: For n=1,LHS=cosα+1-1β=cosαRHS=cosα+1-12βsinβ2sinβ2=cosαAs, LHS=RHSSo, it is true for n=1.Step II: For n=k,Let pk: cosα+cosα+β+cosα+2β+...+cosα+k-1β=cosα+k-12βsinkβ2sinβ2 be true  kN.Step III: For n=k+1,LHS= cosα+cosα+β+cosα+2β+...+cosα+k-1β+cosα+k+1-1β=cosα+k-12βsinkβ2sinβ2+cosα+kβ=cosα+k-12βsinkβ2+sinβ2cosα+kβsinβ2=sinα+kβ-β2-sinα-β2+sinα+kβ+β2-sinα+kβ-β22sinβ2=-sinα-β2+sinα+kβ+β22sinβ2=2cos2α+kβ2sinkβ+β22sinβ2=cosα+kβ2sink+1β2sinβ2RHS=cosα+k+1-12βsink+1β2sinβ2=cosα+kβ2sink+1β2sinβ2As, LHS=RHSSo, it is also true for n=k+1.

Hence, cosα+cosα+β+cosα+2β+...+cosα+n-1β=cosα+n-12βsinnβ2sinβ2 for all nN.

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Question 41:

Prove that 1n+1+1n+2+...+12n>1324, for all natural numbers n>1.                                                                           [NCERT EXEMPLAR]

Answer:

Let pn: 1n+1+1n+2+...+12n>1324, for all natural numbers n>1.Step I: For n=2,LHS=12+1+12×2=13+14=712=1424>1324=RHSAs, LHS>RHSSo, it is true for n=2.Step II: For n=k,Let pk: 1k+1+1k+2+...+12k>1324, be true for some natural numbers k>1.Step III: For n=k+1,pk+1=1k+1+1k+2+...+12k+12k+1>1324+12k+1         Using step II>1324i.e. pk+1>1324So, it is also true for n=k+1.Hence, pn: 1n+1+1n+2+...+12n>1324, for all natural numbers n>1.

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Question 42:

Given a1=12a0+Aa0, a2=12a1+Aa1 and an+1=12an+Aan for n2, where a>0, A>0.Prove that an-Aan+A=a1-Aa1+A2n-1.

Answer:

Given: a1=12a0+Aa0, a2=12a1+Aa1 and an+1=12an+Aan for n2, where a>0, A>0.To prove: an-Aan+A=a1-Aa1+A2n-1Proof:Let pn: an-Aan+A=a1-Aa1+A2n-1Step I: For n=1,LHS=a1-Aa1+ARHS=a1-Aa1+A21-1=a1-Aa1+AAs, LHS=RHSSo, it is true for n=1.Step II: For n=k,Let pk: ak-Aak+A=a1-Aa1+A2k-1 be true for some values of k2.Step III: For n=k+1,pk+1:LHS=ak+1-Aak+1+A=12ak+Aak-A12ak+Aak+A=12ak2+A-2akAak12ak2+A+2akAak=ak2+A-2akAak2+A+2akA=ak-A2ak+A2=ak-Aak+A2=a1-Aa1+A2k-12             Using step II=a1-Aa1+A2k-1×2=a1-Aa1+A2k-1+1=a1-Aa1+A2kRHS=a1-Aa1+A2k+1-1=a1-Aa1+A2kAs, LHS=RHSSo, it is also true for n=k+1.Hence, an-Aan+A=a1-Aa1+A2n-1.

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Question 43:

Let Pn be the statement: 2n3n. If Pr is true, then show that Pr+1 is true. Do you conclude that Pn is true for all nN?

Answer:

Since, for n=1 i.e. P1:LHS=21=2RHS=3×1=3As, LHS<RHSSo, it is not true for n=1.Hence, we conclude that Pn is not true for all nN.

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Question 44:

Show by the Principle of Mathematical induction that the sum Sn of then terms of the series 12+2×22+32+2×42+52+2×62+72+... is given by

Sn=nn+122, if n is evenn2n+12, if n is odd                                                                                                                                         [NCERT EXEMPLAR]

Answer:

Let Pn: Sn=12+2×22+32+2×42+52+...=nn+122, when n is evenn2n+12, when n is oddStep I: For n=1 i.e. P1:LHS=S1=12=1RHS=S1=121+12=1As, LHS=RHSSo, it is true for n=1.Step II: For n=k,Let Pk: Sk=12+2×22+32+2×42+52+...=kk+122, when k is evenk2k+12, when k is odd, be true for some natural numbers.Step III: For n=k+1,Case 1: When k is odd, then k+1 is even.Pk+1:LHS=Sk+1=12+2×22+32+2×42+52+...+k2+2×k+12=k2k+12+2×k+12                  Using step II=k2k+1+4k+122=k+1k2+4k+42=k+1k+222RHS=k+1k+1+122=k+1k+222As, LHS=RHSSo, it is true for n=k+1 when k is odd.Case 2: When k is even, then k+1 is odd.Pk+1:RHS=Sk+1=12+2×22+32+2×42+52+...+2×k2+k+12=kk+122+k+12                  Using step II=kk+12+2k+122=k+12k+22=k+12k+22RHS=k+12k+1+12=k+12k+22As, LHS=RHSSo, it is true for n=k+1 when k is even.Hence, Pn is true for all natural numbers.

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Question 45:

Prove that the number of subsets of a set containing n distinct elements is 2n, for all n  N.                                         [NCERT EXEMPLAR]

Answer:

Let the given statement be defined as Pn: The number of subsets of a set containing n distinct elements=2n, for all nN.Step I: For n=1,LHS= As, the subsets of a set containing only 1 element are: ϕ and the set itself.i.e. the number of subsets of a set containing only 1 element=2RHS=21=2As, LHS=RHSSo, it is true for n=1.Step II: For n=k,Let Pk: The number of subsets of a set containing k distinct elements=2k, be true for some kN.Step III: For n=k+1,Pk+1:Let A=a1, a2, a3, ..., ak, b so that A has k+1 elements.So, the subset of A can be divided into two collections; first contains subsets of A which don't have b in them andthe second contains subsets of A which do have b in them.i.e.First collection: , a1, a1, a2, a1, a2, a3, ..., a1, a2, a3, ..., ak andSecond collection: b, a1, b, a1, a2,, a1, a2, a3, b, ..., a1, a2, a3, ..., ak, bIt can be clearly seen that:The number of subsets of A in first collection=The number of subsets of set with k elements i.e. a1, a2, a3, ..., ak=2k      Using step IIAlso, it follows that the second collection must have the same number of the subsets as that of the first =2kSo, the total number of subsets of A=2k+2k=2×2k=2k+1.

Hence, the number of subsets of a set containing n distinct elements is 2n, for all n  N.

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Question 46:

A sequence a1, a2, a3, ... is defined by letting a1=3 and ak=7ak-1 for all natural numbers k2. Show that an=3·7n-1 for all nN.
                                                                                                                                                                                         [NCERT EXEMPLAR]

Answer:

Let Pn: an=3·7n-1 for all nN.Step I: For n=1,P1:a1=3·71-1=3·1=3So, it is true for n=1.Step II: For n=k,Let Pk: ak=3·7k-1 be true for some kN and k2.Step III: For n=k+1,ak+1=7ak=7·3·7k-1           Using step II=3·7k-1+1=3·7k+1-1So, it is also true for n=k+1.Hence, an=3·7n-1 for all nN.

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Question 47:

A sequence x1, x2, x3, ... is defined by letting x1=2 and xk=xk-1k for all natural numbers k, k2. Show that xn=2n! for all nN.
                                                                                                                                                                                      [NCERT EXEMPLAR]

Answer:

Given: A sequence x1, x2, x3, ... is defined by letting x1=2 and xk=xk-1k for all natural numbers k, k2.Let Pn: xn=2n! for all nN.Step I: For n=1,P1: x1=21!=2So, it is true for n=1.Step II: For n=k,Let Pk: xk=2k! be true for some kN.Step III: For n=k+1,Pk+1:xk+1=xk+1-1k=xkk=2k×k!        Using step II=2k+1!So, it is also true for n=k+1.Hence, xn=2n! for all nN.

Disclaimer: It should be k instead n in the denominator of xk=xk-1k. The same has been corrected above.

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Question 48:

A sequence x0, x1, x2, x3, ... is defined by letting x0=5 and xk=4+xk-1 for all natural number k.Show that xn=5+4n for all nN using mathematical induction.
                                                                                                                                                                                      [NCERT EXEMPLAR]

Answer:

Given: A sequence x0, x1, x2, x3, ... is defined by letting x0=5 and xk=4+xk-1 for all natural number k.Let Pn: xn=5+4n for all nN.Step I: For n=0,P0: x0=5+4×0=5So, it is true for n=0.Step II: For n=k,Let Pk: xk=5+4k be true for some kN.Step III: For n=k+1,Pk+1: xk+1=4+xk+1-1=4+xk=4+5+4k=5+4k+1So, it is also true for n=k+1.Hence, xn=5+4n for all nN.

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Question 49:

Using principle of mathematical induction, prove that n<11+12+13+...+1n for all natural numbers n2.     [NCERT EXEMPLAR]

Answer:

Let Pn: n<11+12+13+...+1n for all natural numbers n2.Step I: For n=2,P2:LHS=21.414RHS=11+12=1+221+0.707=1.707As, LHS<RHSSo, it is true for n=2.Step II: For n=k,Let Pk: k<11+12+13+...+1k be true for some natural numbers n2.Step III: For n=k+1,Pk+1:LHS=k+1RHS=11+12+13+...+1k+1k+1>k+1k+1As, k+1>kkk+1<1kk+1<kk+1k+1-1k+1<kk+1-1k+1<kk+1k+1>k+1i.e. LHS<RHSSo, it is also true for n=k+1.Hence, Pn: n<11+12+13+...+1n for all natural numbers n2.

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Question 50:

The distributive law from algebra states that for all real numbers c, a1 and a2, we have ca1+a2=ca1+ca2.Use this law and mathematical induction to prove that, for all natural numbers, n2, if c, a1, a2, ..., an are any real numbers, thenca1+a2+...+an=ca1+ca2+...+can

Answer:

Given: For all real numbers c, a1 and a2, ca1+a2=ca1+ca2.To prove: For all natural numbers, n2, if c, a1, a2, ..., an are any real numbers, thenca1+a2+...+an=ca1+ca2+...+canProof:Let Pn: ca1+a2+...+an=ca1+ca2+...+can for all natural numbers n2 and c, a1, a2, ..., anR.Step I: For n=2,P2:LHS=ca1+a2RHS=ca1+ca2As, ca1+a2=ca1+ca2           GivenLHS=RHSSo, it is true for n=2.Step II: For n=k,Let Pk: ca1+a2+...+ak=ca1+ca2+...+cak be true for some natural numbers k2 and c, a1, a2, ..., akR.Step III: For n=k+1,Pk+1:LHS=ca1+a2+...+ak+ak+1=ca1+a2+...+ak+ak+1=ca1+a2+...+ak+cak+1=ca1+ca2+...+cak+cak+1         Using step IIRHS=ca1+ca2+...+cak+cak+1As, LHS=RHSSo, it is also true for n=k+1.Hence, for all natural numbers, n2, if c, a1, a2, ..., an are any real numbers, thenca1+a2+...+an=ca1+ca2+...+can.



Page No 12.30:

Question 1:

State the first principle of mathematical induction.

Answer:

Let P(n) be a given statement involving the natural number n such that

(i) The statement is true for n = 1, i.e., P(1) is true (or true for any fixed natural number). This step is known as the Basis step.

(ii) If the statement (called Induction hypothesis) is true for n = k (where k is a particular but arbitrary natural number), then the statement is also true for n = k + 1,

i.e, truth of P(k) implies the truth of P(k + 1). This step is known as the Induction (or Inductive) step.

Then P(n) is true for all natural numbers n.

Note: The first principle of mathematical induction states that if the basis step and the inductive step are proven, then P(nis true for all natural numbers.

Page No 12.30:

Question 2:

Write the set of value of n for which the statement P(n): 2n < n! is true.

Answer:

As, n! > 2n when n > 3.

So, the set of value of n for which the statement P(n): 2n < n! is true = {n  N: n > 3}.

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Question 3:

State the second principle of mathematical induction.

Answer:

Second principle of mathematical induction:

Let P(n) be a given statement involving the natural number n such that

(i) The statement is true for n = 1, i.e., P(1) is true (or true for any fixed natural number). This step is known as the Basis step.

(ii) If the statement (called Induction hypothesis) is true for 1  n  k (where k is a particular but arbitrary natural number), then the statement is also true for n = k + 1,

i.e, truth of P(k) implies the truth of P(k + 1). This step is known as the Induction (or Inductive) step.

Then P(n) is true for all natural numbers n.

Note: The second principle of mathematical induction is completely equivalent to the first principle of mathematical induction which states that if the basis step and the inductive step are proven, then P(n) is true for all natural numbers.

But the only difference is in the inductive hypothesis step 
that we assume not only that the statement holds for n = k but also that it is true for all  n  k.

Also, the base can be other natural number as well apart 1 in both the principles.

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Question 4:

If P(n): 2×42n+1+33n+1 is divisible by λ for all n  N is true, then find the value of λ.

Answer:

For n=1,P1=2×42+1+33+1=2×43+34=128+81=209For n=2,P2=2×44+1+36+1=2×45+37=2048+2187=4235As, HCF209, 4235=11So, 2×42n+1+33n+1 is divisible by 11.Hence, the value of λ is 11.

Page No 12.30:

Question 1:

Make the correct alternative in following question:

If xn - 1 is divisible by x - λ, then the least positive integral value of λ is

(a) 1                    (b) 2                    (c) 3                    (d) 4

Answer:

Let Pn: xn-1 is divisible by x-λ.As, for n=1,P1=x1-1=x-1As, P1 must be divisible by x-λ.x-1 must be divisible by x-λ.So, λ=1

Hence, the correct alternative is option (a).

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Question 2:

Make the correct alternative in the following question:

For all nN, 3×52n+1+23n+1 is divisible by

(a) 19                         (b) 17                         (c) 23                         (d) 25

Answer:

Let Pn= 3×52n+1+23n+1, for all nN.For n=1,P1=3×52+1+23+1=3×53+24=375+16=391=17×23For n=2,P2=3×54+1+26+1=3×55+27=9375+128=9503=17×13×43As, HCF391, 9503=17So, Pn is divisible by 17.

Hence, the correct alternative is option (b).

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Question 3:

Make the correct alternative in the following question:

If 10n+3×4n+2+λ is divisible by 9 for all n N, then the least positive integral value of λ is

(a) 5                         (b) 3                         (c) 7                         (d) 1
 

Answer:

Let Pn: 10n+3×4n+2+λ be divisible by 9 for all nN.For n=1,P1=101+3×41+2+λ=10+3×43+λ=10+192+λ=202+λAs, the least value of P1 which is divisible by 9 is 207.202+λ=207λ=207-202 λ=5

Hence, the correct alternative is option (a).

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Question 4:

Make the correct alternative in the following question:

Let P(n): 2n < (1 × 2 × 3 × ... × n). Then the smallest positive integer for which P(n) is true

(a) 1                         (b) 2                        (c) 3                        (d) 4

Answer:

As, 2n < (1 × 2 × 3 × ... × n) is possible only when n  4

So, the smallest positive integer for which P(n) is true, is 4.

Hence, the correct alternative is option (d).

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Question 5:

Make the correct alternative in the following question:

A student was asked to prove a statement P(n) by induction. He proved P(k +1) is true whenever P(k) is true for all > 5  N and also P(5) is true. On the basis of this he could conclude that P(n) is true.

(a) for all n N                         (b) for all > 5                         (c) for all n 5                         (d) for all n < 5

Answer:

As, P(5) is true and
P(k + 1) is true whenever P(k) is true for all k > 5  N.

By the definition of the priniciple of mathematical induction, we get
P(n) is true for all n  5.

Hence, the correct alternative is option (c).

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Question 6:

Make the correct alternative in the following question:

If P(n): 49n + 16n + λ is divisible by 64 for n  N is true, then the least negative integral value of λ is

(a) -3                         (b) -2                         (c) -1                         (d) -4

Answer:

We have,Pn: 49n+16n+λ is divisible by 64 for all nN.For n=1,P1=491+161+λ=65+λAs, the nearest value of P1 which is divisible by 64 is 64 itself.65+λ=64λ=64-65 λ=-1

Hence, the correct alternative is option (c).



Page No 12.3:

Question 1:

If P (n) is the statement "n(n + 1) is even", then what is P(3)?

Answer:

We have:
P(n): n(n + 1) is even.
Now,
P(3) = 3(3 + 1) = 12       (Even)
Therefore, P(3) is even.

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Question 2:

If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.

Answer:

We have:
P(n): n3+n is divisible by 3.Thus, we have:P(3) =33+3 =27+3=30 It is divisible by 3.Hence, P(3) is true.Now,P(4) =43+4=64+4=68 It is not divisible by 3.Hence, P(4) is not true.

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Question 3:

If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.

Answer:

We have:
P(n): 2n3nAlso,P(r) is true.2r3rTo Prove: P(r+1) is true.We have:2r3r2r×23r×2        Multiplying both sides by 22r+16r2r+13r+3          6r3r+3 for every rN.Hence, P(r+1) is true.

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Question 4:

If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.

Answer:

P(n): n2+n is even.Also, P(r) is true.Thus, r2+r is even.To prove: P(r+1) is true.Now,P(r+1)=(r+1)2+r+1             =r2+1+2r+r+1              =r2+3r+2             =r2+r+2r+2             =P(r)+2(r+1)P(r) is even.Also, 2(r+1) is even, as it is a multiple of 2.Therefore, P(r+1) is even and true.

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Question 5:

Given an example of a statement P (n) such that it is true for all n ∈ N.

Answer:

Proved:
P(n)=n2+n is even for P(n) and P(n+1).Therefore, n2+n 2 is also even for all nN.      Dividing an even number by 2 gives an even number.Thus, we have:P(n) =1+2+...+n          =n(n+1)2         (Even for all nN))

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Question 6:

If P (n) is the statement "n2n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.

Answer:

P(n): n2-n+41 is prime.Now,P(1) =12-1+41=41  (prime)P(2)=22-2+41 =4-2+41 =43  (prime)P(3) =32-3+41=9-3+41=47 (prime)P(41)=412-41+41=1681 (not prime)Thus, we can say that P(1), P(2) and P(3) are true, but P(41) is not true.

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Question 7:

Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.

Answer:

Let P(n) be the statement 3n < n!.

For n = 1,

3n = 3 × 1 = 3

n! = 1! = 1

Now, 3 > 1

So, P(1) is not true.

For n = 2,

3n = 3 × 2 = 6

n! = 2! = 2

Now, 6 > 2

So, P(2) is not true.

For n = 3,

3n = 3 × 3 = 9

n! = 3! = 6

Now, 9 > 6

So, P(3) is not true.

For n = 4,

3n = 3 × 4 = 12

n! = 4! = 24

Now, 12 < 24

So, P(4) is true.

For n = 5,

3n = 3 × 5 = 15

n! = 5! = 120

Now, 15 < 120

So, P(5) is true.

Similarly, it can be verified that 3n < n! for n = 6, 7, 8, ... .

Thus, the statement P(n) : 3n < n! is true for all n ≥ 4 but P(1), P(2) and P(3) are not true.



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