NCERT Solutions For Class 11 Maths chapter-4 Principle of Mathematical Induction

NCERT Solutions For Class 11 Maths Chapter 4 Principle of Mathematical Induction prepared by the expert of Physics Wallah score more with Physics Wallah NCERT Class 11 maths solutions. You can download solution of all chapters from Physics Wallah NCERT solutions of class 11.

NCERT Solutions for Class-11 Maths Chapter 4 Exercise

Prove the following by using the principle of mathematical induction for all n ∈ N:

Question1.1+3+3²+.....+3^n-1=(3ⁿ-1)/2


Solution :
Letp(n):1+3+3²+.....+3^n-1 = (3ⁿ-1)/2

for n = 1

L.H.S = 3^1-1 = 1

Hence by Principle of Mathematical Induction,N is true for all n ∈ N.

 

Question2.

Solution :

Question3.Prove the following by using the principle of mathematical induction for all n∈N .

Solution :

Question4. Prove the following by using the principle of mathematical induction for all n∈ N:

1.2.3 +2.3.4 +…+ n(n + 1)(n + 2) = n(n+1) (n+2) (n+3)/4

Solution :

Therefore, P(k+1) holds whenever P(k)

 holds.

Hence, the given equality is true for all natural numbers i.e., N
 by the principle of mathematical induction.

Question5. Prove the following by using the principle of mathematical induction for all n∈ N:

Solution :

 

Question6. Prove the following by using the principle of mathematical induction for all n∈ N:

Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction.

.

Question7. Prove the following by using the principle of mathematical induction for all n∈ N:

Solution :

Question 8.  Prove the following by using the principle of mathematical induction for all n∈ N:

          1.2+2.2²+3.2²+....+n.2ⁿ=(n-1)2ⁿ⁺¹+2

Solution :
Let p(n) : 1.2+2.2²+3.2²+....+n.2ⁿ=(n-1)2ⁿ⁺¹+2

For n = 1

L.H.S =1.2 = 2

R.H.S.=(1-1)2¹⁺¹+2=0+2=2

Now, let p(n) be true for n = 1

 Let us assume that P(k) is true for some positive integer k, i.e.,

1.2+2.2²+3.2²+...+k.2^k=(k-1)2^k+1+2...(i)

Now, we have to prove that P(k+1) is also true.

Consider

{1.2+2.2²+3.2²+...+k.2^k}+(k+1).^2k+1

=(k-1)^2k+1+2+(k+1)^2k+1

 =2^k+1{(k-1)+(k+1)}+2

 =2^k+1.2k+2

 

=k.2^(k+1)+1 +2

 

={(k+1)-1}2^(k+1)+1+2

=Therefore, P(k+1) holds whenever P(k)

 holds.

Hence, the given equality is true for all natural numbers i.e., N
 by the principle of mathematical induction.
 

Hence by Principle of Mathematical Induction,is true for all n ∈ N.

Question 9. Prove the following by using the principle of mathematical induction for all n∈ N:

 

Question 10. Prove the following by using the principle of mathematical induction for all n∈ N:

Solution :
Let

For n = 1

 

Question11. Prove the following by using the principle of mathematical induction for all n∈ N:

Solution :

Question12. Prove the following by using the principle of mathematical induction for all n∈ N:

Solution :

Question13. Prove the following by using the principle of mathematical induction for all n∈ N:

Solution :

Question14. Prove the following by using the principle of mathematical induction for all n∈ N:

Solution :
Let

Question 15. Prove the following by using the principle of mathematical induction for all n∈ N:

Solution :

 

Question16.

Solution :

  for n = 1

Therefore, P(k+1) holds whenever P(k)holds.

Hence, the given equality is true for all natural numbers i.e., N
 by the principle of mathematical induction.

Question 17. Prove the following by using the principle of mathematical induction for alln∈N:

Solution :

Therefore, P(k+1) holds whenever P(k)holds.

Hence, the given equality is true for all natural numbers i.e., N
 by the principle of mathematical induction.

 

Question18. Prove the following by using the principle of mathematical induction for alln∈N:

Solution :

Question19. Prove the following by using the principle of mathematical induction for alln∈N:n (n + 1) (n + 5) is a multiple of 3.

Solution :Let us denote the given statement by P(n)
 i.e.

P(n):n(n+1)(n+5)  which is a multiple of 3
For n=1
1(1+1)(1+5)=12,
which is a multiple of 3.
Therefore, P(n)
 is true for n=1.
Let us assume that P(k)
 is true for some natural number k,

k(k+1)(k+5)
 is a multiple of 3.
∴k(k+1)(k+5)=3m
, where m∈N
 …(i)

Now, we have to prove that P(k+1)
 is also true whenever P(k)

 is true.

Consider

(k+1){(k+1)+1}{(k+1)+5}

 

=(k+1)(k+2){(k+1)+5}

 

=(k+1)(k+2)(k+5)+(k+1)(k+2)

 

={k(k+1)(k+5)+2(k+1)(k+5)}+(k+1)(k+2)

 

=3m+(k+1){2(k+5)+(k+2)}

 

=3m+(k+1){2k+10+k+2}

 

=3m+(k+1){3k+12}

 

=3m+3(k+1){k+4}

 

=3{m+(k+1)(k+4)}=3 × q
, where q={m+(k+1)(k+4)}

 is some natural number.

Hence, (k+1){(k+1)+1}{(k+1)+5}
 is a multiple of 3

.

Therefore, P(k+1)
 holds whenever P(k)

 holds.

Hence, the given equality is true for all natural numbers i.e., N
 by the principle of mathematical induction.

Question20. Prove the following by using the principle of mathematical induction for alln∈N: 10^2n-1 + 1 is divisible by 11.

Ans.

Let P(n) : 10^2n-1 + 1 is divisible by 11.

For n = 1 is divisible by 11

P(1)=10^2n-1 + 1=11 and P(1) is divisible by 11
Therefore, P(n)
 is true for n=1
Let us assume that P(k)
 is true for some natural number k
 i.e.,

i.e., 10^2n-1 +1
 is divisible by 11

.

∴10^2k-1 +1 = 11m
, where m∈N

…(i)

Now, we have to prove that P(k+1)
 is also true whenever P(k)

 is true.

Consider

10^{2(k+1) -1} + 1

 

=10^2k+2-1 +1

 

=10^2k+1 +1

 

=10^{2(102k-1+1-1)}+1

 

=10²(10^2k-1+1 -1)-10²+1

=10².11m-100+1  Using(i)

=100 × 11m-99

 

=11(100m-9)

 

=11r
, where r=(100m-9)

 is some natural number

Therefore, 10^2(k+1)-1+1
 is divisible by 11

Therefore, P(k+1)
 holds whenever P(k)

 holds.

Hence, the given equality is true for all natural numbers i.e., N
 by the principle of mathematical induction.

Question 21. Prove the following by using the principle of mathematical induction for alln∈N: x²ⁿ –y²n is divisible by x+y.
Solution :

Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction.true.

 

Question22. Prove the following by using the principle of mathematical induction for all n∈N: 3²ⁿ⁺²-8n-9
is divisible by 8.

Solution :
Let 3²ⁿ⁺²-8n-9 is divisible by 8.

For p(n): 3is divisible by 8 = 64 is divisible by 8

=8r, where r=(9m+8k+8) is a natural number
Therefore,
32(k+1)+2-8(k+1)-9
 is divisible by 8
Therefore, P(k+1) holds whenever P(k)
holds.
Hence, the given equality is true for all natural numbers i.e., N
 by the principle of mathematical induction.

Question 23. Prove the following by using the principle of mathematical induction for all n∈N: 41ⁿ – 14ⁿ is a multiple of 27.

Solution :
Let 41ⁿ – 14ⁿ is a multiple of 27.

for  n = 1,

Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction.

 

Question 24.(2n+7) < (n + 3)²

Solution :