Rd Sharma Xi 2018 Solutions for Class 11 Science Math Chapter 21 Some Special Series are provided here with simple step-by-step explanations. These solutions for Some Special Series are extremely popular among Class 11 Science students for Math Some Special Series 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 11 Science Math Chapter 21 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 11 Science Math are prepared by experts and are 100% accurate.

Page No 21.10:

Question 1:

13 + 33 + 53 + 73 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2n-13

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=k=1nTk      = k=1n2k-13      = k=1n8k3-1-6k2k-1      = k=1n8k3-1-12k2+6k      = k=1n8k3-1-12k2+6k      =8k=1nk3-k=1n1-12k=1nk2+6k=1nk      =8n2n+124-n-12nn+12n+16+6 nn+1 2     =2n2n+12-n-2nn+12n+1+3nn+1     =nn+12nn+1-22n+1+3-n     =nn+12n2-2n+1-n     =n2n3-2n2+n+2n2-2n+1-1     =n2n3-n     =n22n2-1

Page No 21.10:

Question 2:

22 + 42 + 62 + 82 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2n2

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=k=1nTk      = k=1n2k2      = k=1n4k2      = 4k=1nk2      = 4nn+12n+16     =2n3n+12n+1

Page No 21.10:

Question 3:

1.2.5 + 2.3.6 + 3.4.7 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=nn+1n+4    =nn2+5n+4    =n3+5n2+4n

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=k=1nk3+5k=1nk2+4k=1nk Sn=k=1nk3+5k=1nk2+4k=1nk Sn=n2n+124+5nn+12n+16+ 4nn+1 2Sn=n2n+124+5nn+12n+16+2nn+1Sn=nn+12nn+12+52n+13+4Sn=nn+12n2+n2+10n+53+4Sn=nn+123n2+3n+20n+10+246Sn=nn+1123n2+23n+34

Page No 21.10:

Question 4:

1.2.4 + 2.3.7 +3.4.10 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=nn+13n+1=n3n2+4n+1=3n3+4n2+n

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=3k=1nk3+4k=1nk2+k=1nk Sn=3k=1nk3+4k=1nk2+k=1nk Sn=3n2n+124+4nn+12n+16+ nn+1 2Sn=3n2n+124+2nn+12n+13+nn+12Sn=nn+123nn+12+42n+13+1Sn=nn+123n2+3n2+8n+43+1Sn=nn+129n2+9n+16n+8+66Sn=nn+1129n2+25n+14

Page No 21.10:

Question 5:

1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:
Tn=1+2+3+4+5+...+n=nn+12=n2+n2

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=k=1nk2+k2Sn=12k=1nk2+kSn=12nn+12n+16+nn+12Sn=nn+142n+13+1Sn=nn+142n+43Sn=nn+12n+412Sn=nn+1n+26

Page No 21.10:

Question 6:

1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=nn+1=n2+n

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=k=1nk2+kSn=k=1nk2+k=1nkSn=nn+12n+16+nn+12Sn=nn+122n+13+1Sn=nn+122n+43Sn=nn+12n+46Sn=nn+1n+23

Page No 21.10:

Question 7:

3 × 12 + 5 ×22 + 7 × 32 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2n+1n2=2n3+n2

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=k=1n2k3+k2Sn=2k=1nk3+k=1nk2Sn=2n2n+124+nn+12n+16Sn=n2n+122+nn+12n+16Sn=nn+12nn+1+2n+13Sn=nn+123n2+3n+2n+13Sn=nn+123n2+5n+13Sn=nn+163n2+5n+1

Page No 21.10:

Question 8:

Find the sum of the series whose nth term is:
(i) 2n2 − 3n + 5
(ii) 2n3 + 3n2 − 1
(iii) n3 − 3n
(iv) n (n + 1) (n + 4)
(v) (2n − 1)2

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

(i)
Tn=2n2-3n+5

Let Sn be the sum of n terms of the given series.

Now,
Sn=k=1nTk 

Sn=k=1n2k2-3k+5Sn=2k=1nk2-3k=1nk + k=1n5Sn=2nn+12n+16-3nn+12+5nSn=2nn+12n+1-9nn+1+30n6Sn=2n2+2n2n+1-9n2-9n+30n6Sn=4n3+4n2+2n2+2n-9n2-9n+30n6Sn=4n3-3n2+23n6Sn=n4n2-3n+236

(ii)
Tn=2n3+3n2-1

Let Sn be the sum of n terms of the given series.

Now,
Sn=k=1nTk 

Sn=k=1n2k3+3k2-1Sn=2k=1nk3+3k=1nk2 - k=1n1Sn=2n2n+124+3nn+12n+16-nSn=n2n+122+nn+12n+12-nSn=n2n+12+nn+12n+1-2n2Sn=n2n2+1+2n+n2+n2n+1-2n2Sn=n4+n2+2n3+2n3+n2+2n2+n-2n2Sn=n4+4n2+4n3-n2Sn=nn3+4n+4n2-12

(iii)
Tn=n3-3n

Let Sn be the sum of n terms of the given series.

Now,
Sn=k=1nTk 

Sn=k=1nk3-3kSn=k=1nk3-k=1n3kSn=n2n+124-3+32+33+34+...+3nSn=n2n+124-33n-13-1Sn=n2n+124-323n-1

(iv)
Tn=nn+1n+4=n2+nn+4=n3+5n2+4n

Let Sn be the sum of n terms of the given series.

Now,
Sn=k=1nTk 

Sn=k=1nk3+5k2+4kSn=k=1nk3 + 5k=1nk2 + 4k=1nkSn=n2n+124+5nn+12n+16+4nn+12Sn=nn+12nn+12+52n+13+4Sn=nn+1123nn+1+102n+1+24Sn=nn+1123n2+23n+34


(v)
Tn=2n-12

Let Sn be the sum of n terms of the given series.

Now,
Sn=k=1nTk 

Sn=k=1n2k-12Sn=k=1n4k2+1-4k Sn=4k=1nk2+1k=1n-4k=1nk Sn=4nn+12n+16+n-4nn+12Sn=nn+1242n+13-4+nSn=nn+128n+4-123+nSn=nn+128n-83+nSn=4nn+1n-13+nSn=n4n+4n-1+3n3Sn=n34n2+4n-4n-4+3Sn=n34n2-1Sn=n32n-12n+1

Page No 21.10:

Question 9:

Find the 20th term and the sum of 20 terms of the series 2 × 4 + 4 × 6 + 6 × 8 + ...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2n2n+2=4n2+4n

For n = 20, we have:

T20=4n2+4n      =4202+420      =1600+80      =1680

Therefore, the 20th term of the given series is 1680.

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1nTk 

Sn=k=1n4k2+4kSn=4k=1nk2+4k=1nk

For n = 20, we have:

S20=4k=120k2+4k=120kS20=42021416+420212S20=40741+4021S20=11480+840 =12320

Hence, the sum of the first 20 terms of the series is 12320.



Page No 21.18:

Question 1:

3 + 5 + 9 + 15 + 23 + ...

Answer:

Let Tn be the nth term and Sn be the sum to n terms of the given series.

Thus, we have:

Sn=3+5+9+15+23+... +Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     3+5+9+15+23+... +Tn-1+Tn   ...(2)

On subtracting (2) from (1), we get:

      Sn    =  3+5+9+15+23+... +Tn-1+Tn      Sn    =        3+5+9+15+23+... +Tn-1+Tn               0      = 3+2+4+6+8+... +Tn-Tn-1-Tn

The sequence of difference of successive terms is 2, 4, 6, 8,...
We observe that it is an AP with common difference 2 and first term 2.

Thus, we have:

3+n-124+n-22-Tn=03+n-122n=Tn3+nn-1=Tn

Now,
 Sn=k=1nTk  Sn=k=1n3+kk-1  Sn=k=1nk2+k=1n3-k=1nkSn=nn+12n+16+3n-nn+12Sn=n3n+12n+12+9-32n+1Sn=nn2+83

Page No 21.18:

Question 2:

2 + 5 + 10 + 17 + 26 + ...

Answer:

Let Tn be the nth term and Sn be the sum of n terms of the given series.

Thus, we have:

Sn=2+5+10+17+26+... +Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     2+5+10+17+26+... +Tn-1+Tn      ...(2)

On subtracting (2) from (1), we get:

Sn= 2+5+10+17+26+... +Tn-1+TnSn=       2+5+10+17+26+... +Tn-1+Tn       0=   2+3+5+7+9+... +Tn-Tn-1-Tn

The sequence of difference of successive terms is 3, 5, 7, 9,...

We observe that it is an AP with common difference 2 and first term 3.

Thus, we have:

2+n-126+n-22-Tn=02+n2-1=Tnn2+1=Tn

Now,
 Sn=k=1nTk  Sn=k=1nk2+1  Sn=k=1nk2+k=1n1Sn=nn+12n+16+nSn=nn+12n+1+6n6Sn=n2n2+3n+76

Page No 21.18:

Question 3:

1 + 3 + 7 + 13 + 21 + ...

Answer:

Let Tn be the nth term and Sn be the sum of n terms of the given series.

Thus, we have:

Sn=1+3+7+13+21+... +Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     1+3+7+13+21+... +Tn-1+Tn       ...(2)

On subtracting (2) from (1), we get:

Sn=1+3+7+13+21+... +Tn-1+TnSn=      1+3+7+13+21+... +Tn-1+Tn0 = 1+2+4+6+8+.. +Tn-Tn-1-Tn=0

The sequence of difference of successive terms is 2, 4, 6, 8,...

We observe that it is an AP with common difference 2 and first term 2.

Thus, we have:

1+n-124+n-22-Tn=01+n2-n=Tnn2-n+1=Tn

Now,
 Sn=k=1nTk  Sn=k=1nk2-k+1  Sn=k=1nk2+k=1n1-k=1nkSn=nn+12n+16+n-nn+12Sn=nn+122n-23+nSn=nn2-1+33Sn=nn2+23

Page No 21.18:

Question 4:

3 + 7 + 14 + 24 + 37 + ...

Answer:

Let Tn be the nth term and Sn be the sum of n terms of the given series.

Thus, we have:

Sn=3+7+14+24+37+ ... +Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     3+7+14+24+37+ ... +Tn-1+Tn      ...(2)

On subtracting (2) from (1), we get:

Sn=3+7+14+24+37+ ... +Tn-1+TnSn=      3+7+14+24+37+ ... +Tn-1+Tn    0  = 3+4+7+10+13+... +Tn-Tn-1-Tn

The sequence of difference of successive terms is 4, 7, 10, 13,...

We observe that it is an AP with common difference 3 and first term 4.

Thus, we have:

3+n-128+n-23-Tn=03+n-123n+2-Tn=03n2-n+42=Tn32n2-n2+2=Tn

Now,
 Sn=k=1nTk  Sn=k=1n32k2-k2+2  Sn=32k=1nk2+k=1n2-12k=1nkSn=nn+12n+14+2n-nn+14Sn=nn+12n+8n4Sn=n+12n2+8n4Sn=n2nn+1+4Sn=n2n2+n+4

Page No 21.18:

Question 5:

1 + 3 + 6 + 10 + 15 + ...

Answer:

Let Tn be the nth term and Sn be the sum of n terms of the given series.

Thus, we have:

Sn=1+3+6+10+15+... +Tn-1+Tn       ....(1)

Equation (1) can be rewritten as:

Sn=     1+3+6+10+15+ ... +Tn-1+Tn            ...(2)

On subtracting (2) from (1), we get:

Sn=1+3+6+10+15+... +Tn-1+TnSn=      1+3+6+10+15+... +Tn-1+Tn 0 =1+2+3+4+5+... +Tn-Tn-1-Tn

The sequence of difference of successive terms is 2, 3, 4, 5,...

We observe that it is an AP with common difference 1 and first term 2.

Thus, we have:

1+n-124+n-21-Tn=01+n-12n+2-Tn=0n2+n2=Tn

Now,
 Sn=k=1nTk  Sn=k=1nk2+k2  Sn=12k=1nk2+12k=1nkSn=nn+12n+112+nn+14Sn=nn+142n+13+1Sn=nn+142n+43Sn=nn+12n+23Sn=nn+1n+26

Page No 21.18:

Question 6:

1 + 4 + 13 + 40 + 121 + ...

Answer:

Let Tn be the nth term and Sn be the sum to n terms of the given series.

Thus, we have:

Sn=1+4+13+40+121+...+Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     1+4+13+40+121+...+Tn-1+Tn      ...(2)

On subtracting (2) from (1), we get:

    Sn=1+4+13+40+121+...+Tn-1+Tn    Sn=      1+4+13+40+121+...+Tn-1+Tn            0 =1+3+9+27+81+... +Tn-Tn-1-Tn

The sequence of difference between successive terms is 3, 9, 27, 81,...

We observe that it is a GP with common ratio 3 and first term 3.

Thus, we have:

1+33n-1-13-1-Tn=01+3n-32-Tn=03n2-12-Tn=03n2-12=Tn

 Sn=k=1nTk  Sn=k=1n3k2-12Sn=12k=1n3k-12k=1n1Sn=123+32+33+34+35+...+3n -n2Sn=1233n-12-n2Sn=3n+1-34-n2Sn=3n+1-3-2n4

Page No 21.18:

Question 7:

4 + 6 + 9 + 13 + 18 + ...

Answer:

Let Tn be the nth term and Sn be the sum of n terms of the given series.

Thus, we have:

Sn=4+6+9+13+18+...+Tn-1+Tn       ...(1)

Equation (1) can be rewritten as:

Sn=     4+6+9+13+18+...+Tn-1+Tn       ...(2)

On subtracting (2) from (1), we get:

    Sn= 4+6+9+13+18+...+Tn-1+Tn    Sn=       4+6+9+13+18+...+Tn-1+Tn    0   =4+2+3+4+5+6+... +Tn-Tn-1-Tn

The sequence of difference between successive terms is 2, 3, 4, 5,...

We observe that it is an AP with common difference 1 and first term 2.

Now,

4+n-124+n-21-Tn=04+n-12n+2-Tn=04+n2+n2-1-Tn=0n22+n2+3=Tn

 Sn=k=1nTk  Sn=k=1nk22+k2+3        =12k=1nk2+12k=1nk+k=1n3        =nn+12n+12×6+nn+12×2+3n        =n2n2+3n+1+3n+3+3612        =n122n2+6n+40        =n6n2+3n+20

Page No 21.18:

Question 8:

2 + 4 + 7 + 11 + 16 + ...

Answer:

Let Sn be the sum of n terms and Tn be the nth term of the given series.

Thus, we have:

Sn=     2+4+7+11+16+...+Tn-1+Tn            ...(1)

Equation (1) can be rewritten as:

Sn=2+4+7+11+16+...+Tn-1+Tn            ...(2)

On subtracting (2) from (1), we get:

    Sn=     2+4+7+11+16+...+Tn-1+Tn    Sn=           2+4+7+11+16+...+Tn-1+Tn    -               -  -  -  -     -            -          -                     0  =   2+2+3+4+5+6+... +Tn-Tn-1-Tn

2+n-124+n-21-Tn=02+n-12n+2-Tn=02+n2+n2-1-Tn=0n22+n2+1=Tn

 Sn=k=1nTk  Sn=k=1nk22+k2+1        =12k=1nk2+12k=1nk+k=1n1        =nn+12n+112+nn+14+n        =n2n2+3n+1+3n+3+1212        =n122n2+6n+16        =n6n2+3n+8

Page No 21.18:

Question 9:

11.4+14.7+17.10+...

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=1(3n-2) (3n+1)

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=k=1n13k-23k+1   =13k=1n13k-2-13k+1   =13k=1n13k-2-13k=1n13k+1   =131+14+17+110+...+13n-2-14+17+110+...+13n-2+13n+1   =131-13n+1   =n3n+1      

Page No 21.18:

Question 10:

11.6+16.11+111.14+114.19+...+1(5n-4) (5n+1)

Answer:

Let Tn be the nth term of the given series.

Thus, we have:

Tn=1(5n-4) (5n+1)

Now, let Sn be the sum of n terms of the given series.

Thus, we have:
Sn=k=1n15k-45k+1    =15k=1n15k-4-15k+1    =15k=1n15k-4-15k=1n15k+1    =151+16+111+116+...+15n-4-16+111+116+...+15n-4+15n+1    =151-15n+1    =n5n+1      

Page No 21.18:

Question 1:

Write the sum of the series 2 + 4 + 6 + 8 + ... + 2n.

Answer:

Sn=2+4+6+8+...+2n 
    =n24+n-12    =n22+2n    =nn+1

Page No 21.18:

Question 2:

Write the sum of the series 12 − 22 + 32 − 42 + 52 − 62 + ... + (2n − 1)2 − (2n)2.

Answer:

The given series can be rewritten as:

3-1 + 7-1 + 11-1 +...+4n-1 -1

=-3+7+11+...+4n-1=-n23×2+n-14=-n24n+2=-n2n+1



Page No 21.19:

Question 3:

Write the sum to n terms of a series whose rth term is r + 2r.

Answer:

Series whose rth term is r + 2r:

1+21+2+22+3+23+4+24+...+n+2n

Thus, we have:

Sn= 1+21+2+22+3+23+4+24+...+n+2n     = 1+2+3+4+...+n+2+22+23+24+...+2n     = nn+12+22n-12-1     = nn+12+2n+1-2

Page No 21.19:

Question 4:

If r=1nr=55, find r=1nr3.

Answer:

r=1nr3 = 13+23+33+...+n3          =nn+122          =r=1nr2          =552          =3025

Page No 21.19:

Question 5:

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then write the value of k.

Answer:

According to the question,
2+4+...+2n = k1+3+5+7+...+2n-1

2×nn+12= kn22×1+n-1×22nn+12= kn22+2n-2nn+1= kn22nn2+n= kn2k=n+1n

Page No 21.19:

Question 6:

Write the sum of 20 terms of the series 1+12(1+2)+13(1+2+3)+....

Answer:

Let the nth term be an.
Here, 
an=1n1+2+3+...+n=n+12
We know:
Sn=k=1nak
Thus, we have:
S20=k=120ak
     =12k=120k+1     =12k=120k+20     =1220212+20     =12230     =115

Page No 21.19:

Question 7:

Write the 50th term of the series 2 + 3 + 6 + 11 + 18 + ...

Answer:

We have,a1=2,a2=3=2+1,a3=6=2+1+3,a4=11=2+1+3+5,...a50=2+1+3+5+...50 terms=2+4922×1+49-1×2             As, the terms apart 2 are in A.P. with a=1 and d=2=2+4922+48×2=2+492×98=2+492=2+2401=2403

So, the 50th term of the given series is 2403.

Page No 21.19:

Question 8:

Let Sn denote the sum of the cubes of first n natural numbers and sn denote the sum of first n natural numbers. Then, write the value of
r=1nSrsr.

Answer:

We know that, Sr=13+23+33+...+r3=rr+122And, sr=1+2+3+...+r=rr+12As, Srsr=rr+122rr+12=rr+12=12r2+rNow,r=1nSrsr=r=1n12r2+r=12r=1nr2+r=1nr=12nn+12n+16+nn+12=12×nn+12×2n+13+1=nn+142n+1+33=nn+142n+43=nn+14×2n+23=nn+1n+26

Page No 21.19:

Question 1:

The sum to n terms of the series 11+3+13+5+15+7+....+.... is
(a) 2n+1

(b) 122n+1

(c) 2n+1-1

(d) 122n+1-1

Answer:

(d) 122n+1-1

Let Tn be the nth term of the given series.

Thus, we have:
Tn=12n-1+2n+1=2n+1-2n-12

Now,

Let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=k=1nTk      = k=1n2k+1-2k-12     =12 k=1n2k+1-2k-1     = 123-1+5-3+7-5+...+2n+1-2n-1     = 12-1+2n+1     = 122n+1-1

Page No 21.19:

Question 2:

The sum of the series 1log2 4+1log4 4+1log8 4+....+1log2n4 is
(a) n (n+1)2

(b) n (n+1) (2n+1)12

(c) n (n+1)4

(d) none of these

Answer:

(c) n (n+1)4

Let Sn=1log2 4+1log4 4+1log8 4+...+1log2n4 

Sn=log2log4+log4log4+log8log4+...+log2nlog4Sn=log2log4+log22log4+log23log4+...+log2nlog4  Sn=log2log4+2 log2log4+3 log2log4+...+n log2log4  Sn=log2log41+2+3+...+n Sn=log412log41+2+3+...+n Sn=12log4log41+2+3+...+n Sn=121+2+3+...+n Sn=nn+14

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Question 3:

The value of r=1n(2r-1) a+1br is equal to

(a) an2+bn-1-1bn-1 (b-1)

(b) an2+bn-1bn (b-1)

(c) an3+bn-1-1bn (b-1)

(d) none of these

Answer:

(b) an2+bn-1bn (b-1)

We have:
r=1n(2r-1) a+1br=r=1n2ra-a+1br=r=1n2ar-r=1na+r=1n1br=ann+1-an+1-bn1-bbn=an2+bn-1b-1bn

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Question 4:

If ∑ n = 210, then ∑ n2 =
(a) 2870
(b) 2160
(c) 2970
(d) none of these

Answer:

(a) 2870

Given:
n = 210

nn+12=210n2+n-420=0n-20n+21=0n=20                          n>0

Now,
n2=nn+12n+16n(n+1)2×(2n+1)3210 × 41370 × 412870

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Question 5:

If Sn = r=1n1+2+22+... Sum to r terms2r, then Sn is equal to
(a) 2nn − 1

(b) 1-12n

(c) n-1+12n

(d) 2n − 1

Answer:

(c) n-1+12n

We have:
Sn = r=1n1+2+22+...sum to r terms2r
Sn=r=1n12r-12rSn=r=1n1-12rSn=n-r=1n12rSn=n- 121 - 12n 1-12Sn=n- 1-12nSn=n-1+12n



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Question 6:

If 1+1+22+1+2+33+.... to n terms is S, then S is equal to

(a) n (n+3)4

(b) n (n+2)4

(c) n (n+1) (n+2)6

(d) n2

Answer:

a n (n+3)4

Let Tn be the nth term of the given series.

Thus, we have:

Tn=1+2+3+4+5+...+nn=nn+12n=n2+12

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn= k=1nk2+12 Sn= k=1nk2+n2 Sn= nn+14+n2 Sn= n2n+12+1 Sn= n2n+32 Sn= nn+34

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Question 7:

Sum of n terms of the series 2+8+18+32+ .... is
(a) n (n+1)2

(b) 2n (n + 1)

(c) n (n+1)2

(d) 1

Answer:

(c) n (n+1)2

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2×n2=n2

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=2 k=1nk Sn= 2nn+12 Sn= nn+12

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Question 8:

The sum of 10 terms of the series 2+6+18+.... is
(a) 121 (6+2)

(b) 243 (3+1)

(c) 1213-1

(d) 242 (3-1)

Answer:

(a) 121 (6+2)

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2×3n-1=23n-1

Now, let S10 be the sum of 10 terms of the given series.

Thus, we have:

 S10=2 k=1103k-1 S10= 21+3+32+...+39 S10= 2310-13-1 S10n= 235-13-13+13+1 S10= 2235-13+1 S10= 122426+2 S10= 121 6+2

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Question 9:

The sum of the series 12 + 32 + 52 + ... to n terms is
(a) n (n+1) (2n+1)2

(b) n (2n-1) (2n +1)3

(c) (n-1)2 (2n+1)6

(d) (2n+1)33

Answer:

(b) n (2n-1) (2n +1)3

Let Tn be the nth term of the given series.

Thus, we have:

Tn=2n-12=4n2+1-4n

Now, let Sn be the sum of n terms of the given series.

Thus, we have:

Sn= k=1n4k2+1-4k Sn= 4   k=1   nk2 + 1k=1n - 4k=1nk Sn= 4nn+12n+1 6+n- 4nn+12 Sn= 2nn+12n+1 3+n- 2nn+1 Sn= n2n+12n+1 3+1- 2n+1 Sn= n32n+22n+1 +3 -6n+1 Sn= n34n2-1 Sn= n2n-12n+1 3

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Question 10:

The sum of the series 23+89+2627+8081+..... to n terms is
(a) n-12(3-n-1)

(b)n-12(1-3-n)

(c) n+12(3n-1)

(d) n-12(3n-1)

Answer:

(b) n-12(1-3-n)

Let Tn be the nth term of the given series.

Thus, we have:

Tn=3n-13n=1-13n

Now,

Let Sn be the sum of n terms of the given series.

Thus, we have:

Sn=k=1nTk      = k=1n1-13k     = k=1n1- k=1n13k     = n-13+132+133+...+13n     = n-131-13n1-13     = n-121-13n     = n-121-3-n



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