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Satvik Kamat
Subject: Maths
, asked on 8/9/17
For all positive integers 𝑛, prove that n
7
/7 + n
5
/5 + 2𝑛3/3 − n/105 is an integer.
Answer
2
Jayesh Kumar
Subject: Maths
, asked on 31/8/17
Prove by mathematical induction that the product of three consective natural numbers is always divisible by 6.
Answer
1
Chharishma.r.nayaka
Subject: Maths
, asked on 22/8/17
If 'a' and 'b' are natural numbers such that a
2
-b
2
is a prime number then a
2
-b
2
equals
(a) a+b
(b) a-b
(c) ab
(d) 1
Answer
1
Chharishma.r.nayaka
Subject: Maths
, asked on 22/8/17
If a,b and n are natural numbers then a
2n-1
+ b
2n-1
is divisible by
(a) a-b
(b) a
3
+b
3
(c) a+b
(d) a
2
+b
2
Answer
1
Chharishma.r.nayaka
Subject: Maths
, asked on 22/8/17
Sum to n terms of the series 1 + (1+x) + (1+x+x2) + (1+x+x2+x3) + ....
Answer
2
Kiran
Subject: Maths
, asked on 20/8/17
Que:Prove by the principal of mathematica induction that :n(n+1)(2n+1) is dived by 6 for all nEN.
Hint: R.D.Sharma(12.4 example 4) please explain in form of first principal of mathematical induction
Answer
1
Anwesha Biswal
Subject: Maths
, asked on 19/8/17
4^x1=5 , 5^x2=6 , 6^x3=7, ..... , 126^x123=127 , 127^x124=128. what is the value of product x1 x2 x3......x124
Answer
1
Parameswaran Namboothiri
Subject: Maths
, asked on 18/8/17
prove by pmi
Answer
1
Jasnoor Kaur
Subject: Maths
, asked on 15/8/17
Let p(n) :2^n
Answer
1
Royel Ghost
Subject: Maths
, asked on 14/8/17
Please explain the answer.
Example 6 prove that
2.7
n
+ 3.5
n
– 5 is divisible by 24, for all
n
∈
N.
solution Let the statement p (
n
) be defined as
P(
n
) : 2.7
n
+3.5
n
– 5 is divisible by 24.
We note that
P(n
) is true for
n =
1, since 2.7 + 3.5 5 = 24, which is divisible by 24.
Assume that p(
k
) is true
i.e 2.7
k
+3.5
k
– 5 = 24
q
, when
q
∈
N
Now we wish to prove that P(
k
+ 1) is true whenever P(
k
) is true.
We have
2
.
7
k
+
1
+
3
.
5
k
+
1
–
5
=
2
.
7
k
.
7
l
+
3
.
5
k
.
5
l
–
5
=
7
[
2
.
7
k
+
3
.
5
k
–
5
–
3
.
5
k
+
5
]
+
3
.
5
k
.
5
–
5
=
7
[
24
q
–
3
.
5
k
+
5
]
+
15
.
5
k
–
5
=
7
×
24
q
–
21
.
5
k
+
35
+
15
.
5
k
–
5
=
7
×
24
q
–
6
.
5
k
+
30
=
7
×
24
q
–
6
(
5
k
–
5
)
=
7
×
24
q
–
6
(
4
p
)
[
(
5
k
–
5
)
is
a
multiple
of
4
(
why
?
)
]
=
7
×
24
q
–
24
p
=
24
(
7
q
–
p
)
=
24
×
r
,
r
=
7
q
–
p
,
is
some
natural
number
.
.
.
.
(
2
)
The expression on the R.H.S. of (1) is divisible by 24. Thus P(
k
+ 1) is true whenever P(
k
) true
Hence, by principle of mathematical induction, P(
n
) is true for all
n
∈
N.
Answer
1
Amitesh Kumar
Subject: Maths
, asked on 10/8/17
Solve this:
Answer
1
Lioncub
Subject: Maths
, asked on 29/7/17
Prove by using PMI:
1+5+9+..+(4n-3) = n(2n-1) for all natural numbers n
Answer
3
Lioncub
Subject: Maths
, asked on 29/7/17
Prove using PMI:
1+2+22+...+2^n = 2^(n+1) -1 for all natural numbers n.
Answer
1
Arjun Singh
Subject: Maths
, asked on 28/7/17
1+2+3+4+...+n=1/2n(n+1) induction
Answer
1
Ananya Pandey
Subject: Maths
, asked on 25/7/17
Questions 4, 5,6
Answer
1
Prev
2
3
4
5
6
Next
What are you looking for?
(a) a+b
(b) a-b
(c) ab
(d) 1
(a) a-b
(b) a3+b3
(c) a+b
(d) a2+b2
Hint: R.D.Sharma(12.4 example 4) please explain in form of first principal of mathematical induction
Example 6 prove that
2.7n+ 3.5n – 5 is divisible by 24, for all n N.
solution Let the statement p (n) be defined as
P(n) : 2.7n +3.5n – 5 is divisible by 24.
We note that P(n) is true for n =1, since 2.7 + 3.5 5 = 24, which is divisible by 24.
Assume that p(k) is true
i.e 2.7k+3.5k – 5 = 24q, when q N
Now we wish to prove that P(k + 1) is true whenever P(k) is true.
We have
The expression on the R.H.S. of (1) is divisible by 24. Thus P(k + 1) is true whenever P(k) true
Hence, by principle of mathematical induction, P(n) is true for all n N.
1+5+9+..+(4n-3) = n(2n-1) for all natural numbers n
1+2+22+...+2^n = 2^(n+1) -1 for all natural numbers n.
1+2+3+4+...+n=1/2n(n+1) induction