NCERT Solutions for Class 11 Maths

Class 11 NCERT Solutions- Chapter 4 Principle of Mathematical Induction – Exercise 4.1 | Set 1

Last Updated : 29 Nov, 2022

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

Question 1: 1 + 3 + 3² + …….. + 3^n-1 = $\frac{3^n - 1}{2}$

Solution:

We have,
P(n) = 1 + 3 + 3² + …….. + 3^n-1 = $\frac{3^n - 1}{2}$
For n=1, we get
P(1) = 1 = $\frac{3^1 - 1}{2} = \frac{3-1}{2} = 1$
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = 1 + 3 + 3² + …….. + 3^k-1 = $\frac{3^k - 1}{2}$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = 1 + 3 + 3² + …….. + 3^k-1 + 3^(k+1)-1
= (1 + 3 + 3² + …….. + 3^k-1) + 3^k
From Eq(1), we get
= $\frac{3^k - 1}{2}$ + 3^k
= $\frac{3^k - 1 + 2(3^k)}{2}$
= $\frac{3.3^k - 1}{2}$
= $\frac{3^{k+1} - 1}{2}$
Hence,
P(k+1) = $\frac{3^{k+1} - 1}{2}$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 2: 1 + 2³ + 3³ + ……….. + n³ = $[\frac{n(n+1)}{2} ]^2$

Solution:

We have,
P(n) = 1 + 2³ + 3³ + ……….. + n³ = $[\frac{n(n+1)}{2} ]^2$
For n=1, we get
P(1) = 1 = $[\frac{1(1+1)}{2} ]^2$ = 1
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = 1 + 2³ + 3³ + ……….. + k³ = $[\frac{k(k+1)}{2} ]^2$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = 1 + 2³ + 3³ + ……….. + k³ + (k+1)³
= (1 + 2³ + 3³ + ……….. + k³) + (k+1)³
From Eq(1), we get
= $[\frac{k(k+1)}{2} ]^2$ + (k+1)³
= $\frac{k^2(k+1)^2}{4}$ + (k+1)³
= $\frac{k^2(k+1)^2 + 4(k+1)^3}{4}$
Taking (k+1)², we get
= $\frac{(k+1)^2(k^2 + 4(k+1))}{4}$
= $\frac{(k+1)^2(k^2 + 4k + 4)}{4}$
= $\frac{(k+1)^2(k+2)^2}{2^2}$
= $[\frac{(k+1)((k+1)+1)}{2} ]^2$
Hence,
P(k+1) = $[\frac{(k+1)((k+1)+1)}{2} ]^2$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 3: 1 + $\frac{1}{(1+2)} + \frac{1}{(1+2+3)}$ + ……. + $\frac{1}{(1+2+3+....n)}$ = $\frac{2n}{(n+1)}$

Solution:

We have,
P(n) = 1 + $\frac{1}{(1+2)} + \frac{1}{(1+2+3)} + ....... + \frac{1}{(1+2+3+....n)}$ = $\frac{2n}{(n+1)}$
For n=1, we get
P(1) = 1 = $\frac{2(1)}{1+1}$ = 1
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = 1 + $\frac{1}{(1+2)} + \frac{1}{(1+2+3)} + ....... + \frac{1}{(1+2+3+....k)}$ = $\frac{2k}{(k+1)}$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = 1 + $\frac{1}{(1+2)} + \frac{1}{(1+2+3)} + ....... + \frac{1}{(1+2+3+....k)}$ + $\frac{1}{(1+2+3+....k+(k+1))}$
= (1 + $\frac{1}{(1+2)} + \frac{1}{(1+2+3)} + ....... + \frac{1}{(1+2+3+....k)}$ ) + $\frac{1}{(1+2+3+....k+(k+1))}$
From Eq(1), we get
= $\frac{2k}{(k+1)} + \frac{1}{(1+2+3+....k+(k+1))}$
As we know that,
Sum of first natural number,
1 + 2 + 3 + …… + n = $\frac{n(n+1)}{2}$
So, we get
= $\frac{2k}{(k+1)} + \frac{1}{\frac{(k+1)(k+1+1)}{2}}$
= $\frac{2k}{(k+1)} + \frac{2}{(k+1)(k+2)}$
= $\frac{2}{(k+1)} (k+\frac{1}{k+2})$
= $\frac{2}{(k+1)} (\frac{(k(k+2) + 1)}{k+2})$
= $\frac{2}{(k+1)} (\frac{(k^2+2k+1)}{k+2})$
= $\frac{2}{(k+1)} (\frac{(k+1)^2}{k+2})$
= $\frac{2(k+1)}{(k+1)+1}$
Hence,
P(k+1) = $\frac{2(k+1)}{(k+1)+1}$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

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

Solution:

We have,
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 get
P(1) = 1.2.3 = $\frac{1(1+1)(1+2)(1+3)}{4}$ = $\frac{1.2.3.4}{4}$ = 1.2.3
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = 1.2.3 + 2.3.4 +…+ k(k+1) (k+2) = $\frac{k(k+1)(k+2)(k+3)}{4}$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = 1.2.3 + 2.3.4 +…+ k(k+1) (k+2) + (k+1)(k+1+1) (k+1+2)
= (1.2.3 + 2.3.4 +…+ k(k+1) (k+2)) + (k+1)(k+2) (k+3)
From Eq(1), we get
= $\frac{k(k+1)(k+2)(k+3)}{4}$ + (k+1)(k+2)(k+3)
= (k+1)(k+2) (k+3) ( $\frac{k}{4}$ + 1)
= $(k+1)(k+2) (k+3) (\frac{k + 4}{4})$
= $\frac{(k+1)((k+1)+1) ((k+1)+2) ((k+1)+3)}{4}$
Hence,
P(k+1) = $\frac{(k+1)((k+1)+1) ((k+1)+2) ((k+1)+3)}{4}$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 5: 1.3 + 2.3² + 3.3³ +…+ n.3ⁿ = $\frac{(2n-1)3^{n+1}+3}{4}$

Solution:

We have,
P(n) = 1.3 + 2.3² + 3.3³ +…+ n.3ⁿ = $\frac{(2n-1)3^{n+1}+3}{4}$
For n=1, we get
P(1) = 1.3 = 3 = $\frac{(2(1)-1)3^{1+1}+3}{4} = \frac{9+3}{4} = \frac{12}{4}$ = 3
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = 1.3 + 2.3² + 3.3³ +…+ k.3^k = $\frac{(2k-1)3^{k+1}+3}{4}$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = 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)
From Eq(1), we get
= $\frac{(2k-1)3^{k+1}+3}{4}$ + (k+1).3^(k+1)
= $\frac{(2k-1)3^{k+1}+3 + 4(k+1).3^{k+1}}{4}$
= 3^k+1 $\frac{((2k-1) + 4(k+1)) + 3}{4}$
= 3^k+1 $\frac{(6k+3) + 3}{4}$
= 3^k+1 $\frac{(3(2k+1)) + 3}{4}$
= 3^(k+1)+1 $\frac{(2(k+1)-1) + 3}{4}$
Hence,
P(k+1) = 3^(k+1)+1 $\frac{(2(k+1)-1) + 3}{4}$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 6: 1.2 + 2.3 + 3.4 +…+ n.(n+1) = $[\frac{n(n+1)(n+2)}{3}]$

Solution:

We have,
P(n) = 1.2 + 2.3 + 3.4 +…+ n.(n+1) = $[\frac{n(n+1)(n+2)}{3}]$
For n=1, we get
P(1) = 1.2 = 2 = $[\frac{1(1+1)(1+2)}{3}] = [\frac{1(2)(3)}{3}]$ = 2
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = 1.2 + 2.3 + 3.4 +…+ k.(k+1) = $[\frac{k(k+1)(k+2)}{3}]$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = 1.2 + 2.3 + 3.4 +…+ k.(k+1) + (k+1)(k+1+1)
= (1.2 + 2.3 + 3.4 +…+ k.(k+1)) + (k+1)(k+2)
From Eq(1), we get
= $[\frac{k(k+1)(k+2)}{3}]$ + (k+1)(k+2)
= $[\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}]$
= (k+1)(k+2) $[\frac{k+3}{3}]$
= $[\frac{(k+1)(k+2)(k+3)}{3}]$
= $[\frac{(k+1)((k+1)+1)((k+1)+2)}{3}]$
Hence,
P(k+1) = $[\frac{(k+1)((k+1)+1)((k+1)+2)}{3}]$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 7: 1.3 + 3.5 + 5.7 +…+ (2n–1) (2n+1) = $\frac{n(4n^2+6n-1)}{3}$

Solution:

We have,
P(n) = 1.3 + 3.5 + 5.7 +…+ (2n–1) (2n+1) = $\frac{n(4n^2+6n-1)}{3}$
For n=1, we get
P(1) = 1.3 = 3 = $\frac{1(4(1)^2+6(1)-1)}{3}$ = $\frac{9}{3}$ = 3
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = 1.3 + 3.5 + 5.7 +…+ (2k–1) (2k+1) = $\frac{k(4k^2+6k-1)}{3}$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = 1.3 + 3.5 + 5.7 +…+ (2k–1) (2k+1) + (2(k+1)-1)(2(k+1)+1)
= (1.3 + 3.5 + 5.7 +…+ (2k–1) (2k+1)) + (2k+1)(2k+3)
From Eq(1), we get
= $\frac{k(4k^2+6k-1)}{3}$ + (4k²+8k+3)
= $\frac{k(4k^2+6k-1)+3(4k^2+8k+3)}{3}$
= $\frac{4k^3+6k^2-k+12k^2+24k+9}{3}$
= $\frac{4k^3+18k^2+23k+9}{3}$
= $\frac{4k^3+14k^2+9k+4k^2+14k+9}{3}$
= $\frac{k(4k^2+14k+9)+1(4k^2+14k+9)}{3}$
= $\frac{(k+1) (4k^2+14k+9)}{3}$
= $\frac{(k+1) (4(k^2+2k+1)+6(k+1)-1)}{3}$
= $\frac{(k+1) (4(k+1)^2+6(k+1)-1)}{3}$
Hence,
P(k+1) = $\frac{(k+1) (4(k+1)^2+6(k+1)-1)}{3}$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 8: 1.2 + 2.2² + 3.2³ + …+n.2ⁿ = (n–1) 2^{n + 1} + 2

Solution:

We have,
P(n) = 1.2 + 2.2² + 3.2³ + …+n.2ⁿ = (n–1) 2^{n + 1} + 2
For n=1, we get
P(1) = 1.2 = 2 = (1–1) 2(1) + 1 + 2 = 2
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = 1.2 + 2.2² + 3.2³ + …+k.2^k = (k–1) 2^{k + 1} + 2 ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = 1.2 + 2.2² + 3.2³ + …+k.2^k + (k+1).2^(k+1)
= (1.2 + 2.2² + 3.2³ + …+k.2^k) + (k+1).2^(k+1)
From Eq(1), we get
= (k–1) 2^{k + 1} + 2 + (k+1).2^k+1
= 2^{k + 1}((k–1) + (k+1)) + 2
= 2k + 1(2k) + 2
= k.2^k+1+1 + 2
= ((k+1)-1).2^(k+1)+1 + 2
Hence,
P(k+1) = ((k+1)-1).2^(k+1)+1 + 2
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 9: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8}$ + …… + $\frac{1}{2^n} = 1 - \frac{1}{2^n}$

Solution:

We have,
P(n) = $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...... + \frac{1}{2^n} = 1 - \frac{1}{2^n}$
For n=1, we get
P(1) = $\frac{1}{2}$ = 1 – $\frac{1}{2^1}$ = $\frac{1}{2}$
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...... + \frac{1}{2^k} = 1 - \frac{1}{2^k}$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...... + \frac{1}{2^k} + \frac{1}{2^{k+1}}$
= ( $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...... + \frac{1}{2^k}$ ) + $\frac{1}{2^{k+1}}$
From Eq(1), we get
= 1 – $\frac{1}{2^k} + \frac{1}{2^{k+1}}$
= 1 – $\frac{1}{2^k} + \frac{1}{2.2^k}$
= 1 – $\frac{1}{2^k}(1-\frac{1}{2})$
= 1 – $\frac{1}{2^k}(\frac{1}{2})$
= 1 – $\frac{1}{2^{k+1}}$
Hence,
P(k+1) = 1 – $\frac{1}{2^{k+1}}$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 10: $\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11}$ + …… + $\frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}$

Solution:

We have,
P(n) = $\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ...... + \frac{1}{(3n-1)(3n+2)} = \frac{n}{(6n+4)}$
For n=1, we get
P(1) = $\frac{1}{2.5} = \frac{1}{10} = \frac{1}{(6(1)+4)} = \frac{1}{10}$
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = $\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ...... + \frac{1}{(3k-1)(3k+2)} = \frac{k}{(6k+4)}$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = $\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ...... + \frac{1}{(3k-1)(3k+2)} + \frac{1}{(3(k+1)-1)(3(k+1)+2)}$
= ( $\frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} + ...... + \frac{1}{(3k-1)(3k+2)}$ ) + $\frac{1}{(3k+2)(3k+5)}$
From Eq(1), we get
= $\frac{k}{(6k+4)} + \frac{1}{(3k+2)(3k+5)}$
= $\frac{k}{2(3k+2)} + \frac{1}{(3k+2)(3k+5)}$
= $\frac{1}{3k+2} (\frac{k}{2} + \frac{1}{3k+5})$
= $\frac{1}{3k+2} (\frac{k(3k+5)+1(2)}{2(3k+5)})$
= $\frac{1}{3k+2} (\frac{3k^2+5k+2}{6k+10})$
= $\frac{1}{3k+2} (\frac{(3k+2)(k+1)}{6k+10})$
= $\frac{(k+1)}{6(k+1)+4}$
Hence,
P(k+1) = $\frac{(k+1)}{6(k+1)+4}$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 11: $\frac{1}{1.2.3} + \frac{1}{2.3.5} + \frac{1}{3.4.5}$ + …… + $\frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$

Solution:

We have,
P(n) = $\frac{1}{1.2.3} + \frac{1}{2.3.5} + \frac{1}{3.4.5}+ ...... + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$
For n=1, we get
P(1) = $\frac{1}{1.2.3} = \frac{1}{6} = \frac{1(1+3)}{4(1+1)(1+2)} = \frac{4}{4(2)(3)} = \frac{1}{6}$
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = $\frac{1}{1.2.3} + \frac{1}{2.3.5} + \frac{1}{3.4.5}+ ...... + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)}$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = $\frac{1}{1.2.3} + \frac{1}{2.3.5} + \frac{1}{3.4.5}+ ...... + \frac{1}{k(k+1)(k+2)} + \frac{1}{(k+1)(k+1+1)(k+1+2)}$
= ( $\frac{1}{1.2.3} + \frac{1}{2.3.5} + \frac{1}{3.4.5}+ ...... + \frac{1}{k(k+1)(k+2)}$ ) + $\frac{1}{(k+1)(k+2)(k+3)}$
From Eq(1), we get
= $\frac{1}{(k+1)(k+2)} (\frac{k(k+3)}{4} + \frac{1}{(k+3)})$
= $\frac{1}{(k+1)(k+2)} (\frac{k(k+3)^2 + 4}{4(k+3)})$
= $\frac{1}{(k+1)(k+2)} (\frac{k(k^2+6k+9) + 4}{4(k+3)})$
= $\frac{1}{(k+1)(k+2)} (\frac{k^3+6k^2+9k + 4}{4(k+3)})$
= $\frac{1}{(k+1)(k+2)} (\frac{k^3+2k^2+k + 4k^2 + 8k+ 4}{4(k+3)})$
= $\frac{1}{(k+1)(k+2)} (\frac{k(k^2+2k+1) + 4(k^2 + 2k+ 1)}{4(k+3)})$
= $\frac{1}{(k+1)(k+2)} (\frac{(k+4)(k^2+2k+1)}{4(k+3)})$
= $\frac{1}{(k+1)(k+2)} (\frac{(k+4)(k+1)^2}{4(k+3)})$
= $\frac{1}{(k+2)} (\frac{(k+4)(k+1)}{4(k+3)})$
= $\frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2))}$
Hence,
P(k+1) = $\frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2))}$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 12: a + ar + ar² + …… + ar^n-1 = $\frac{a(r^n-1)}{r-1}$

Solution:

We have,
P(n) = a + ar + ar² + …… + ar^n-1 = $\frac{a(r^n-1)}{r-1}$
For n=1, we get
P(1) = a = $\frac{a(r^1-1)}{r-1} = \frac{a(r-1)}{r-1}$ = a
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = a + ar + ar² + …… + ar^k-1 = $\frac{a(r^k-1)}{r-1}$ ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = a + ar + ar² + …… + ar^k-1 + ar^(k+1)-1
= (a + ar + ar² + …… + ar^k-1) + ar^k
From Eq(1), we get
= $\frac{a(r^k-1)}{r-1}$ + ar^k
= $\frac{a(r^k-1) + ar^k(r-1)}{r-1}$
= $\frac{a(r^k-1 + r^{k+1}-r^k)}{r-1}$
= $\frac{a(r^{k+1}-1)}{r-1}$
Hence,
P(k+1) = $\frac{a(r^{k+1}-1)}{r-1}$
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.

Question 13: (1+ $\frac{3}{1}$ ) (1+ $\frac{5}{4}$ ) (1+ $\frac{7}{9}$ ) ….. (1+ $\frac{(2n+1)}{n^2}$ ) = (n+1)²

Solution:

We have,
P(n) = $(1+ \frac{3}{1}) (1+ \frac{5}{4}) (1+ \frac{7}{9}) ..... (1+\frac{(2n+1)}{n^2})$ = (n+1)²
For n=1, we get
P(1) = 1+ $\frac{3}{1}$ = 1+3 = 4 = (1+1)² = 2² = 4
So, P(1) is true
Assume that P(k) is true for some positive integer n=k
P(k) = $(1+ \frac{3}{1}) (1+ \frac{5}{4}) (1+ \frac{7}{9}) ..... (1+\frac{(2k+1)}{k^2})$ = (k+1)² ……………..(1)
Let’s prove that P(k + 1) is also true. Now, we have
P(k+1) = $(1+ \frac{3}{1}) (1+ \frac{5}{4}) (1+ \frac{7}{9}) ..... (1+\frac{(2k+1)}{k^2}) (1+\frac{(2(k+1)+1)}{(k+1)^2})$
= $((1+ \frac{3}{1}) (1+ \frac{5}{4}) (1+ \frac{7}{9}) ..... (1+\frac{(2k+1)}{k^2})) (1+\frac{2(k+1)+1}{(k+1)^2})$
From Eq(1), we get
= (k+1)² (1+ $\frac{2(k+1)+1}{(k+1)^2}$ )
= (k+1)² $(\frac{(k+1)^2 + 2(k+1)+1}{(k+1)^2})$
= (k+1)² + 2(k+1) + 1
= {(k+1)}²
Hence,
P(k+1) = {(k+1)}²
Thus, P(k + 1) is true, whenever P (k) is true.
Hence, from the principle of mathematical induction, the statement P(n) is true for all natural numbers n.