OP Malhotra Class 11 Maths Solutions Chapter 8 Mathematical Induction Ex 8(a)

Continuous practice using S Chand Class 11 Maths Solutions Chapter 8 Mathematical Induction Ex 8(a) can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 8 Mathematical Induction Ex 8(a)

Use the principle of mathematical induction to prove the following statements for all n ∈ N.

Question 1.
1 + 2 + 3 + … + n = \(\frac { 1 }{ 2 }\) n(n + 1).
Solution:
Let T (n) be the statement: 1 + 2 + 3 + … + n = \(\frac { 1 }{ 2 }\) (n +1)
For n = 1 ; 1 = \(\frac { 1 }{ 2 }\) x 1 (1 +1) which is true. ∴ T (1) is true.
Now let us assume that T (k) is true
i.e. 1 + 2 + 3 + … + k = \(\frac { 1 }{ 2 }\) k (k +1)
Now, we shall prove that T (k + 1) is true.
Now, 1 + 2 + 3 + … + k + k + 1 = \(\frac { 1 }{ 2 }\) k (k + 1) + k + 1 [using (1)]
= \(\frac { k+1 }{ 2 }\)[k + 2] = \(\frac { 1 }{ 2 }\)(k + 1)(k + 1 + 1)
∴ T(k + 1) is true.
Hence by mathematical induction T (n) is true for all n ∈ N.

Question 2.
2 + 4 + 6 +… + 2n = n (n + 1).
Solution:
For n = 1 ; 2 = 1 (1 + 1)
∴ result is true for n = 1
Let us assume that result is true for n = m
i.e. 2 + 4 + 6 + … + 2m = m (m + 1) … (1)
We shall prove the result for n = m + 1
i.e. we want to prove that
2 + 4 + 6 + …. + 2 (m – 1) = (m + 1) (m + 2)
Now 2 + 4 + 6 + …. + 2m + 2 {m + 1) = m (m + 1) + 2 (m + 1) [using eqn. (1)]
= (m + 1) (m + 2)
= (m + 1) (m + 1 + 1)
∴ result is true for n = m + 1
Hence by mathematical induction result is true for all n ∈ N.

Question 3.
1² + 2² + 3² +…+ n² = \(\frac { 1 }{ 6 }\)n (n + 1) (2n + 1).
Solution:
For n = 1 ; 1² = \(\frac { 1 }{ 6 }\) 1 (1 + 1) (2.1 + 1) = \(\frac { 6 }{ 6 }\) = 1
∴ result is true for n = 1
Let us assume that result is true for n = m
∴ 1² + 2² + 3² + …. + m² = \(\frac { 1 }{ 6 }\) m.(m + 1)(2m + 1) … (1)
We shall prove the result for n = m + 1

∴ result is true for n = m + 1
Hence by principle of mathematical induction result is true for all n ∈ N.

Question 4.
1² + 3² + 5² +… + (2n – 1)² = \(\frac { 1 }{ 3 }\) n (2n – 1) (2n + 1)
Solution:
Let P (n) be the statement:
1² + 3² + 5² + … + (2n – 1)² = \(\frac { 1 }{ 3 }\) n (2n – 1) (2n + 1)
For n = 1 ; 1² = \(\frac { 1 }{ 3 }\) x 1 (2 – 1)(2 + 1) = 1
∴ P (1) is true
Let us assume that P (m) is true, where m ∈ N
i.e. 1² + 3² + 5² …. + (2m – 1)² = \(\frac { 1 }{ 3 }\) m(2m – 1)(2m + 1) … (1)
Now we shall prove that P (m + 1) is true.
Now P (m + 1): 1² + 3² + 5² … + (2m – 1)² + (2m + 1)²
= \(\frac { m }{ 3 }\) (2m – 1) (2m + 1) + (2m + 1)² [using (1)]
= \(\frac{(2 m+1)}{3}\) [m (2m – 1) + 3 (2m +1)] = \(\frac { 1 }{ 3 }\) (2m + 1) (2m² + 5m + 3)
= \(\frac { 1 }{ 3 }\) (2m + 1) (m + 1) (2m + 3) = \(\frac { 1 }{ 3 }\) (m + 1) (2 (m + 1) – 1) (2(m + 1) + 1)
Thus P (m + 1) is true
Hence by mathematical induction result is true for all n ∈ N.

Question 5.
2 + 2² + 2³ + … + 2ⁿ = 2 (2n – 1).
Solution:
Let P (n) be the given statement: 2 + 2² + 2³ + … + 2ⁿ = 2 (2ⁿ – 1)
For n = 1 ; 2 = 2 (2¹ – 1) = 2 x 1 = 2 ∴ P(1) is true
Let us assume that P (m) is true, where m ∈ N
i.e. 2 + 2² + 2³ + … + 2^m = 2 (2^m – 1) … (1)
We shall prove that P (m + 1) is true.
Now P (m + 1): 2 + 2² + 2³ + … + 2^m + 2^m+1
= (2 + 2² + 2³ … + 2^m) + 2^m+1
= 2(2^m – 1) + 2^m+1 [using eqn. (1)]
= 2^m+1 – 2 + 2^m+1 = 2 . 2^m+1 – 2
= 2 (2^m+1 – 1)
Thus P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N

Question 6.
1.2 + 2.3 + 3.4 + … + n (n + 1) = \(\frac{n(n+1)(n+2)}{3}\).
Solution:
Let P (n) be the statement:
1.2 + 2.3 + 3.4 + … + n (n + 1) = \(\frac{n(n+1)(n+2)}{3}\)
For n = 1 ; 1.2 = \(\frac{1(1+1)(1+2)}{3}\) = 2 ∴ P (1) is true.
Let us assume that P (n) is true for n = m ∈ N
i.e. 1.2 + 2.3 + … + m (m + 1) = \(\frac{m(m+1)(m+2)}{3}\) … (1)
We shall prove that P (m+ 1) is true.
Now P (m + 1): 1.2 + 2.3 + 3.4 + … + m (m + 1) + (m + 1) (m + 2)
= [1.2 + 2.3 + … + m (m + 1)] + (m + 1) (m + 2)
= \(\frac{m(m+1)(m+2)}{3}\) + (m + 1) (m + 2) [using eqn. (1)]
= \(\frac{(m+1)(m+2)(m+3)}{3}=\frac{(m+1)(m+1+1)(m+1+2)}{3}\)
∴ P(m + 1) is true.
Hence by mathematical induction result P (n) is true for all n ∈ N.

Question 7.
1.2 + 2.2² + 3.2³ +… + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2.
Solution:
Let P (n) be the statement : 1.2 + 2.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2
For n = 1 : 1.2 = (1 – 1)2¹⁺¹ + 2 = 2 ∴ P(1) is true.
Let us assume that P (m) is true where m ∈ N
i.e. 1.2 + 2.2² + 3.2² + … m . 2^m = (m – 1) 2^m+1 + 2 …(1)
Now we shall prove that P (m + 1) is true
Now P (m + 1) : 1.2 + 2.2² + 3.2² + …. + m.2^m + (m + 1) 2^m+1
= (1.2 + 2.2² + 3.2³ + … + m.2^m) + (m + 1) 2^m+1
= (m – 1) 2^m+1 + 2 + (m + 1) 2^m+1 [using (1)]
= (m – 1 + m + 1). 2^m+1 + 2 = 2m. 2^m+1 + 2 = (m + 1 – 1) 2^m+2 + 2
∴ P (m + 1) is true.
Hence by principle of mathematical induction P (n) is true or all n ∈ N.

Question 8.
3.2² + 3².2³ + 3³.2⁴ + … + 3ⁿ . 2ⁿ⁺¹ = \(\frac { 12 }{ 5 }\)(6ⁿ – 1).
Solution:
Let P (n) be the statement: 3.2² + 3².2³ + 3³.2⁴ + … + 3ⁿ . 2ⁿ⁺¹ = \(\frac { 12 }{ 5 }\)(6ⁿ – 1)
For n = 1 ; 3.2² = \(\frac { 12 }{ 5 }\) (6¹ – 1) = 12
∴ P (1) is true.
Let us assume that P (m) is true.
i.e. 3.2 + 3².2³ + … + 3^m.2^m+1 = \(\frac { 12 }{ 5 }\) (6^m – 1) … (1)
Now we shall prove that P (m + 1) is true.

∴ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

Question 9.
1.3 + 3.5 + 5.7 +… + (2n – 1) (2n + 1) = \(\frac{n\left(4 n^2+6 n-1\right)}{3}\).
Solution:
Let P (n) be the statement: 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = \(\frac{n\left(4 n^2+6 n-1\right)}{3}\)
For n = 1; 1.3 = \(\frac{1 \cdot\left(4 \cdot 1^2+6 \cdot 1-1\right)}{3}\) = 3 ∴ P(1) is true
Let us assume that P (m) is true, where m ∈ N
i.e 1.3 + 3.4 + … + (2m – 1) (2m + 1) = \(\frac{m\left(4 m^2+6 m-1\right)}{3}\)
Now we shall prove that P (m + 1) is true.

∴ P (m + 1) is true
Hence by mathematical induction P (n) is true for all n ∈ N.

Question 10.
(i) a + (a + d) + (a + 2d) +… + [a + (n -1) d] = \(\frac { n }{ 2 }\) [2a + (n – 1) d]
(ii) a + ar + ar² +… + ar^n-1 = \(\frac{a\left(1-r^n\right)}{1-r}\), r ≠ 1.
Solution:
(i) Let P (n) be the statement: a + (a + d) + (a + 2d) + … + [a + (n – 1) d] = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
For n = 1 ; a = \(\frac { 1 }{ 2 }\) [2a + (1 – 1)d] = a ∴ P (1) is true.
Let us assume that P (m) is true where m ∈ N
i.e. a + (a + d) (a + 2d) + …. + [a + (m – 1) d] = \(\frac { m }{ 2 }\) [2a + (m – 1) d] …(1)
Now, we shall prove that ∴ P (m + 1) is true.
Now P (m + 1) ; a + (a + d) + (a + 2d) + … + [a + (m – 1) d] + [a + (m + 1 – 1) d]
= a + (a + d)(a + 2d) + … + [a + (m – 1) d] + (a + md)
= \(\frac { m }{ 2 }\) [2a + (m – 1)d] + (a + md) [using eqn. (1)]
= (ma + a) + \(\frac { md }{ 2 }\) [(m – 1) + 2]
= (m + 1)a + \(\frac { m }{ 2 }\)(m + 1)
= \(\frac { (m+1) }{ 2 }\) [2a + (m + 1 – 1)d]
∴ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

(ii) Let P (n) be the statement: a + ar + ar² + … + ar^n-1 = \(\frac{a\left(1-r^n\right)}{1-r} ; r \neq 1\)
For n = 1 ; a = \(\frac{a\left(1-r^1\right)}{1-r}\) = a ∴ P (1) is true.
Let us assume that P (m) is true where m ∈ N

Hence by principle of mathematical induction P (n) is true for all n ∈ N.

Question 11.
5 + 15 + 45 + … + 5 (3)^n-1 = \(\frac { 5 }{ 2 }\) (3ⁿ – 1).
Solution:
Let P (n) be the statement: 5 + 15 + 45 + … + 5 (3)^n-1 = \(\frac { 5 }{ 2 }\) (3ⁿ – 1)
For n = 1 ; 5 = \(\frac { 5 }{ 2 }\) (3¹ – 1) ∴ P(1) is true
Let us assume that P (m) is true where m ∈ N
i.e. 5 + 15 + 45 + … + 5.3^m-1 = \(\frac { 5 }{ 2 }\) (3^m – 1) …(1)
We shall prove that ∴ P (m + 1) is true.
Now P (m + 1): 5 + 15 + 45 + … + 5.3^m-1 + 5.3^m+1-1
= [5 + 15 + 45 + …. + 5.3^m-1] + 5.3^m
= \(\frac { 5 }{ 2 }\) (3^m – 1) + 5.3^m [using eqn. (1)]
= \(\frac{5}{2}\left[3^m-1+2 \cdot 3^m\right]=\frac{5}{2}\left[3 \cdot 3^m-1\right]=\frac{5}{2}\left[3^{m+1}-1\right]\)
∴ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

Question 12.
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}\).
Solution:
Let P(n) be the statement : \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}\)
For n = 1 ; \(\frac{1}{3.5}=\frac{1}{3(2.1+3)}=\frac{1}{15}\) ∴ P (1) is true.

Question 13.
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}\).
Solution:

Question 14.
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^n}=1-\frac{1}{2^n}\).
Solution:

Question 15.
1 + 5 + 12 + 22 + 35 + … + \(\frac { n }{ 2 }\) (3n – 1) = \(\frac{n^2(n+1)}{2}\).
Solution:
Let P (n) be the statement: 1 + 5 + 12 + 22 + … + \(\frac { n }{ 2 }\)(3n – 1) = \(\frac{n^2(n+1)}{2}\)
For n = 1; 1 = \(\frac{1^2(1+1)}{2}\) ∴ P(1) is true.
Let us assume that P (m) is true, where m ∈ N
i.e. 1 + 5 + 12 + 22 + … + \(\frac { m }{ 2 }\) (3m – 1) = \(\frac{m^2(m+1)}{2}\) … (1)
Now we shall prove that P (m + 1) is true.

Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 16.
Prove by the method of mathematical induction that
\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\ldots+\frac{1}{(3 n+1)(3 n+4)}=\frac{n}{4(3 n+4)}\), for n ∈ N.
Solution:

Question 17.
Let S (k) = 1 + 3 + 5 +… + (2k – 1) = 3 + k², then which of the following is true ?
(a) S (k) = S (k + 1)
(b) S(k) ≠ S(k + 1)
(c) S (1) is correct
(d) Principle of mathematical induction can be used to prove the formula.
Solution:
Given S (k) = 1 + 3 + 5 …. + (2k – 1) = 3 + k²
Here S (1); 1 ≠ 4 – 3 + 1²
∴ S (1) is not true.
Thus option (d) is not correct as basic step of mathematical induction is not true.
∴ S(k + 1)= 1 +3 + 5 + … + [2(k + 1) – 1]
= 1 + 3 + 5 + … + (2k + 1)
= (1 + 3 + 5 + … + 2k – 1) + 2k + 1
= 3 + k² + 2k + 1
= 3 + (k + 1)²
∴ S(k) ≠ S(k+ 1)