Question

NCERT Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction is one of the important chapters of mathematics. These NCERT Solutions study materials have a significant weightage in CBSE board exams.

One must refer to our NCERT Solutions Class 11 to gain complete clarity of difficult concepts of the subject. It is the best way to prepare for the board exams.

• Principle of Mathematical Induction – mathematical induction is a way of proving a statement, equation, theorem, or formula which is supposed to be true for every term. Generalizing the mathematical induction for every term will help prove any mathematical statement.

• Process of the proof by induction – in any given statement or the equation m(n). we can use the below method to prove the principle of mathematical induction.
  • Step 1: check for the given statement is true for n = 1.
  • Step 2: take an assumption of m(n) is true for n = k, where k is a positive integer.
  • Step 3: now prove that the result is also true for the next term m(n+1) for any positive integer k.
  • In the above steps if it is found to be true for every term then it is true for all terms.  

• Motivation – in mathematical induction motivation is termed as the proof of the given statement of n natural number in such a way that it holds for all the terms if it is true for the first term.
• Simple applications of mathematical induction – the application of mathematical induction is used to find the next term of any series by proving it true for the first term and subsequent terms as well.

In NCERT Solutions Class 11 Maths all the concepts are discussed in stepwise form with the help of equations, graphs, shortcuts, tips, and tricks. As suggested by our experts, one should refer to our solutions to gain complete clarity of tough questions and equations.

Answer 1

NCERT Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction

1. Prove the following by using the principle of mathematical induction for all n \(\in\) N:

1 + 3 + 3² +... + 3^n-1 = \(\frac{(3^n-1)}{2}\) 

Answer:

Let the given statement be P(n), i.e., 

P(n): 1 + 3 + 3² + …+ 3^n–1 = \(\frac{(3^n-1)}{2}\) 

For n = 1, we have 

P(1):= \(\frac{(3^1-1)}{2}=\frac{3-1}{2}=\frac22=1\) , which is true.

Let P(k) be true for some positive integer k, i.e.,

 1 + 3 + 3² + …+ 3^k–1 = \(\frac{(3^k-1)}{2}\)      ......(i)

We shall now prove that P(k + 1) is true. 

Consider

1 + 3 + 3² + … + 3^k–1 + 3^{(k+1) – 1} 

= (1 + 3 + 3² +… + 3^k–1) + 3^k

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

2. Prove the following by using the principle of mathematical induction for all n \(\in\) N:

1³ + 2³ + 3³ + .... + n³ = \(\left(\frac{n(n+1)}{2}\right)^2\)

Answer:

Let the given statement be P(n), i.e.,

P(n):  1³ + 2³ + 3³ + .... + n³ = \(\left(\frac{n(n+1)}{2}\right)^2\) 

For n = 1, we have

P(1): 1³ = 1 = \(\left(\frac{1(1+1)}{2}\right)^2\) = \(\left(\frac{1.2}{2}\right)^2\) = 1² = 1, which is true.

Let P(k) be true for some positive integer k, i.e.,

1³ + 2³ + 3³ + .... + k³ = \(\left(\frac{k(k+1)}{2}\right)^2\)     .....(i)

We shall now prove that P(k + 1) is true. 

Consider

1³ + 2³ + 3³ + … + k³ + (k + 1)³ 

= (1³ + 2³ + 3³ + …. + k³) + (k + 1)³

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

3. Prove the following by using the principle of mathematical induction for all n \(\in\) N:

Answer:

Let the given statement be P(n), i.e.,

P(n):

For n = 1, we have 

P(1): 1 = \(\frac{2.1}{1+1}=\frac22=1\) , which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

4. Prove the following by using the principle of mathematical induction for all n \(\in\) N: 

1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)

Answer:

Let the given statement be P(n), i.e.,

P(n): 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)

For n = 1, we have 

P(1): 1.2.3 = 6 = \(\frac{1(1+1)(1+2)(1+3)}{4}\) , which is true. 

Let P(k) be true for some positive integer k, i.e.,

1.2.3 + 2.3.4 + … + k(k + 1) (k + 2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\)  ....(i)

We shall now prove that P(k + 1) is true. 

Consider 

1.2.3 + 2.3.4 + … + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3) 

= {1.2.3 + 2.3.4 + … + k(k + 1) (k + 2)} + (k + 1) (k + 2) (k + 3)

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

5. Prove the following by using the principle of mathematical induction for all n \(\in\) N:

1.3 + 2.3² + 3.3³ + ... + n.3ⁿ = \(\frac{(2n-1)3^{n+1}+3}{4}\)

Answer:

Let the given statement be P(n), i.e.,

P(n) :  1.3 + 2.3² + 3.3³ + ... + n.3ⁿ = \(\frac{(2n-1)3^{n+1}+3}{4}\)

For n = 1, we have

P(1): 1.3 = 3 = \(\frac{(2.1-1)3^{1+1}+3}{4}=\frac{3^2+3}{4}=\frac{12}4=3\) , which is true. 

Let P(k) be true for some positive integer k, i.e.,

 1.3 + 2.3² + 3.3³ + ... + k.3^k = \(\frac{(2k-1)3^{k+1}+3}{4}\)   ...(i)

We shall now prove that P(k + 1) is true. 

Consider

1.3 + 2.3² + 3.3³ + … + k.3^k + (k + 1).3^k+1 

= (1.3 + 2.3² + 3.3³ + …+ k.3^k) + (k + 1).3^k+1

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

Answer 2

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

1.2 + 2.3 + 3.4 + ... + n.(n+1) = \(\left[\frac{n(n+1)(n+2)}{3}\right]\)

Answer:

Let the given statement be P(n), i.e.,

P(n):  1.2 + 2.3 + 3.4 + ... + n.(n+1) = \(\left[\frac{n(n+1)(n+2)}{3}\right]\)

For n = 1, we have

P(1): 1.2 = 2 = \(\frac{1(1+1)(1+2)}{3}\) = \(\frac{1.2.3}{3} = 2\) , which is true. 

Let P(k) be true for some positive integer k, i.e.,

 1.2 + 2.3 + 3.4 + ... + k.(k+1) = \(\left[\frac{k(k+1)(k+2)}{3}\right]\)     ....(i)

We shall now prove that P(k + 1) is true. 

Consider 

1.2 + 2.3 + 3.4 + … + k.(k + 1) + (k + 1).(k + 2) 

= [1.2 + 2.3 + 3.4 + … + k.(k + 1)] + (k + 1).(k + 2)

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

1.3 + 3.5 + 5.7 + ....+ (2n - 1) (2n + 1) = \(\frac{n{(4n^2 + 6n -1)}}{3}\)

Answer:

Let the given statement be P(n), i.e., 

P(n):  1.3 + 3.5 + 5.7 + ....+ (2n - 1) (2n + 1) = \(\frac{n{(4n^2 + 6n -1)}}{3}\)

For n = 1, we have

P(1) : 1.3 = 3 = \(\frac{1{(4.1^2 + 6n -1)}}{3}=\frac{4+6-1}{3}=\frac93=3\) , which is true. 

Let P(k) be true for some positive integer k, i.e.,

 1.3 + 3.5 + 5.7 + ....+ (2k - 1) (2k + 1) = \(\frac{k{(4k^2 + 6k -1)}}{3}\)      .....(i)

We shall now prove that P(k + 1) is true. 

Consider

(1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + {2(k + 1) – 1}{2(k + 1) + 1}

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

Answer:

Let the given statement be P(n), i.e., 

P(n): 1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2 

For n = 1, we have

P(1): 1.2 = 2 = (1 – 1) 2¹⁺¹ + 2 = 0 + 2 = 2, which is true. 

Let P(k) be true for some positive integer k, i.e., 

1.2 + 2.2² + 3.2² + … + k.2^k = (k – 1) 2^{k + 1} + 2 … (i)

We shall now prove that P(k + 1) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

\(\frac12+\frac14+\frac18+...+\frac1{2^n}=1-\frac{1}{2^n}\)

Answer:

Let the given statement be P(n), i.e.,

P(n): \(\frac12+\frac14+\frac18+...+\frac1{2^n}=1-\frac{1}{2^n}\)

For n = 1, we have 

P(1): \(\frac12=1-\frac1{2^1}=\frac12\) , which is true. 

Let P(k) be true for some positive integer k, i.e.,

\(\frac12+\frac14+\frac18+...+\frac1{2^k}=1-\frac{1}{2^k}\)      ..... (i)

We shall now prove that P(k + 1) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

Answer:

Let the given statement be P(n), i.e.,

P(n):

For n = 1, we have

P(1) = \(\frac{1}{2.5}=\frac1{10}=\frac{1}{6.1+4}=\frac{1}{10}\) , which is true. 

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}\)

Answer:

Let the given statement be P(n), i.e., 

P(n): \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}\) 

For n = 1, we have

P(1) : \(\frac{1}{1.2.3}=\frac{1.(1+3)}{4(1+1)(1+2)}=\frac{1.4}{4.2.3}=\frac{1}{1.2.3}\) , which is true. 

Let P(k) be true for some positive integer k, i.e.,

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)}\)     ......(i)

We shall now prove that P(k + 1) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

Answer 3

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

a + ar + ar² + .... + ar^n-1 = \(\frac{a(r^n-1)}{r-1}\)

Answer:

Let the given statement be P(n), i.e.,

P(n):  a + ar + ar² + .... + ar^n-1 = \(\frac{a(r^n-1)}{r-1}\)

For n = 1, we have

P(1) : a = \(\frac{a(r^n-1)}{r-1}\) = a , which is true. 

Let P(k) be true for some positive integer k, i.e.,

 a + ar + ar² + .... + ar^k-1 = \(\frac{a(r^k-1)}{r-1}\)     ......(i)

We shall now prove that P(k + 1) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

Answer:

Let the given statement be P(n), i.e.,

For n = 1, we have

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

\((1+\frac11)(1+\frac12)(1+\frac13)....(1+\frac1n)= (n+1)\)

Answer:

Let the given statement be P(n), i.e.,

P(n): \((1+\frac11)(1+\frac12)(1+\frac13)....(1+\frac1n)= (n+1)\)

For n = 1, we have

P(1): \((1+\frac11)=2=(1+1)\) , which is true. 

Let P(k) be true for some positive integer k, i.e.,

P(k): \((1+\frac11)(1+\frac12)(1+\frac13)....(1+\frac1k)= (k+1)\)      .....(1)

We shall now prove that P(k + 1) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

1² + 3² + 5² + ... + (2n - 1)² = \(\frac{n(2n-1)(2n+1)}{3}\) 

Answer:

Let the given statement be P(n), i.e.,

P(n) = 1² + 3² + 5² + ... + (2n - 1)² = \(\frac{n(2n-1)(2n+1)}{3}\) 

For n = 1, we have

P(1) = 12 = 1 = \(\frac{1(2.1-1)(2.1+1)}{3}= \frac{1.1.3}{3}=1\) , which is true.

Let P(k) be true for some positive integer k, i.e.,

P(k) = 1² + 3² + 5² + ... + (2k - 1)² = \(\frac{k(2k-1)(2k+1)}{3}\)      ......(1)

We shall now prove that P(k + 1) is true. Consider

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

\(\frac1{1.4}+\frac1{4.7}+\frac1{7.10}+....+\frac1{(3n-2)(3n+1)}=\frac n{(3n+1)}\)

Answer:

Let the given statement be P(n), i.e.,

P(n): \(\frac1{1.4}+\frac1{4.7}+\frac1{7.10}+....+\frac1{(3n-2)(3n+1)}=\frac n{(3n+1)}\) 

For n = 1, we have 

P(1) = \(\frac{1}{1.4}=\frac{1}{3.1+1}=\frac{1}{4}=\frac{1}{1.4}\) , which is true.

Let P(k) be true for some positive integer k, i.e.,

P(k) = \(\frac1{1.4}+\frac1{4.7}+\frac1{7.10}+....+\frac1{(3k-2)(3k+1)}=\frac k{(3k+1)}\)    .....(1)

We shall now prove that P(k + 1) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}\)

Answer:

Let the given statement be P(n), i.e.,

P(n) = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n}{3(2n+3)}\)  

For n = 1, we have

P(1): \(\frac{1}{3.5} = \frac{1}{3(2.1+3)}=\frac{1}{3.5}\) , which is true. 

Let P(k) be true for some positive integer k, i.e.,

P(k): \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2k+1)(2k+3)}=\frac{k}{3(2k+3)}\)      .....(1)

We shall now prove that P(k + 1) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

Answer 4

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

1 + 2 + 3 + .... + n < \(\frac18\) (2n + 1)²

Answer:

Let the given statement be P(n), i.e.,

P(n): 1 + 2 + 3 + .... + n < \(\frac18\) (2n + 1)²

It can be noted that P(n) is true for n = 1 since

1 < \(\frac18\) (2.1 + 1)² = \(\frac98\) .

Let P(k) be true for some positive integer k, i.e.,

1 + 2 + 3 + .... + k < \(\frac18\) (2k + 1)²    .....(1)

We shall now prove that P(k + 1) is true whenever P(k) is true. 

Consider

Hence, (1 + 2 + 3 +...+ k) + (k + 1)  < \(\frac18\) (2k + 1)² + (k + 1)

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

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

Answer:

Let the given statement be P(n), i.e., 

P(n): n (n + 1) (n + 5), which is a multiple of 3. 

It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3.

Let P(k) be true for some positive integer k, i.e., 

k (k + 1) (k + 5) is a multiple of 3. 

∴ k (k + 1) (k + 5) = 3m, where m ∈ N … (1) 

We shall now prove that P(k + 1) is true whenever P(k) is true. 

Consider

is some natural number 

Therefore, (k + 1) {(k + 1) +1} {(k + 1) + 5} is multiple of 3.

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

10^{2n – 1} + 1 is divisible by 11.

Answer:

Let the given statement be P(n), i.e., 

P(n): 10^{2n – 1} + 1 is divisible by 11. 

It can be observed that P(n) is true for n = 1 

since P(1) = 10^{2.1 – 1} + 1 = 11, which is divisible by 11.

Let P(k) be true for some positive integer k, 

i.e., 10^{2k – 1} + 1 is divisible by 11.

∴ 10^{2k – 1} + 1 = 11m, where m ∈ N … (1) 

We shall now prove that P(k + 1) is true whenever P(k) is true. Consider

= 1 lr, where r = (100m - 9) is some natural number

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

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

x²ⁿ – y²ⁿ is divisible by x + y.

Answer:

Let the given statement be P(n), i.e., 

P(n): x²ⁿ – y²ⁿ is divisible by x + y. 

It can be observed that P(n) is true for n = 1.

This is so because x^{2 × 1} – y^{2 × 1} = x² – y² = (x + y) (x – y) is divisible by (x + y). 

Let P(k) be true for some positive integer k, i.e.,

x^2k – y^2k is divisible by x + y. 

∴ Let x^2k – y^2k = m (x + y), where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true. 

Consider

which is a factor of (x + y).

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

22. Prove the following by using the principle of mathematical induction for all n ∈ N: 3^{2n + 2} – 8n – 9 is divisible by 8.

Answer:

Let the given statement be P(n), i.e., 

P(n): 3^{2n + 2} – 8n – 9 is divisible by 8. 

It can be observed that P(n) is true for n = 1

since 3^{2 × 1 + 2} – 8 × 1 – 9 = 64, which is divisible by 8.

Let P(k) be true for some positive integer k, 

i.e., 3^{2k + 2} – 8k – 9 is divisible by 8. 

∴ 3^{2k + 2} – 8k – 9 = 8m; where m ∈ N … (1) 

We shall now prove that P(k + 1) is true whenever P(k) is true. 

Consider

= 8r, where r = (9m + 8k +8) is a natural number 

Therefore, 3^2(k+1)+2 - 8 (k + 1) -9 is divisible by 8.

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

Answer 5

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

41ⁿ – 14ⁿ is a multiple of 27.

Answer:

Let the given statement be P(n), i.e.,

P(n):41ⁿ – 14ⁿ is a multiple of 27. 

It can be observed that P(n) is true for n = 1

since 41¹ - 14¹ = 27, which is a multiple of 27. 

Let P(k) be true for some positive integer k, i.e.,

41^k – 14^k is a multiple of 27 

∴ 41^k – 14^k = 27m, where m ∈ N ……………. (1) 

We shall now prove that P(k + 1) is true whenever P(k) is true. 

Consider

= 27 x r, where r = (41m - 14k) is a natural number

Therefore, 41^k+1 - 14^k+1 is a multiple of 27.

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.

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

(2n +7) < (n + 3)²

Answer:

Let the given statement be P(n), i.e., 

P(n): (2n +7) < (n + 3)² 

It can be observed that P(n) is true for n = 1 

since 2.1 + 7 = 9 < (1 + 3)² = 16, which is true. 

Let P(k) be true for some positive integer k, i.e.,

(2k + 7) < (k + 3)² … (1) 

We shall now prove that P(k + 1) is true whenever P(k) is true. 

Consider

Thus, P(k + 1) is true whenever P(k) is true. 

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.