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

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

Using the principle of mathematical induction prove that:

Question 1.
n (n + 1) (n + 5) is a multiple of 6.
Solution:
Let P (n) be the statement: n (n + 1) in + 5) is a multiple of 6.
For n = 1 ; 1 (1 + 1) (1 + 6) i.e. 12 is multiple of 6, which is true ∴ P (1) is true.
Let us assume that P (m) is true, where m ∈ N
i.e. m (m + 1) (m + 5) is a multiple of 6
⇒ m (m + 1) (m + 5) = 6k, where k ∈ N …(1)
We shall prove that P (m + 1) is true.
Now P (m + 1): (m + 1) (m + 1 + 1) (m + 1 + 5) = (m + 1) (m + 2) (m + 6)
= (m + 1) [m² + 8m + 12] = (m + 1) [m² + 5m + 3m+ 12]
= (m + 1) (m² + 5m) + (m + 1) 3 (m + 4)
= (m + 1) m (m + 5) + 3m (m + 1) + 12 (m + 1)
= 6k + 3.2k’ + 12 (m + 1) [∵ product of n positive integers is divisible by n!]
= 6 [k + k’ + 2 (m + 1)] = 6k”, which is divisible by 6
where k” = k + k’ + 2 (m + 1) ∈ N
∴ P (m + 1) is true.
Hence by mathematical induction P (m) is true for all n ∈ N

Question 2.
n³ + (n + 1)³ + (n + 2)³ is a multiple of 9.
Solution:
Let P (n) be the statement: n³ + (n + 1)³ + (n + 2)³ is a multiple of 9.
For n = 1 ; 1³ + (1 + 1)³ + (1 + 2)³ = 1 + 8 + 27 = 36, which is a multiple of 9.
∴ P(1) is true.
Let us assume that P (m) is true where m ∈ N,
i.e. m³ + (m + 1 )³ + (m + 2)² is a multiple of 9.
∴ m³ + (m + 1)³ + (m + 2)³ = 9k …(1)
Now we shall prove that P (m + 1) is true.
Now P (m + 1) : (m + 1)³ + (m + 2)³ + (m + 3)³
= (m + 1)³ + (m + 2)³ + m³ + 27 + 9m (m + 3)
= 9k + 9m (m + 3) + 27 [using eqn. (1)]
= 9λ, where λ = k + m(m + 3) + 3 ∈ N which is a multiple of 9.
∴ P (m + 1) is true.
Hence by mathematical induction P (m) is true for all n ∈ N.

Question 3.
2^{3n – 1} is divisible by 7.
Solution:
Let P (n) be the statement: 2^{3n – 1} is divisible by 7.
For n = 1 ; 2^3×1 – 1 = 8 – 1 = 7, which is divisible by 7 ∴ P (1) is true.
Let us assume that P (m) is true, where m ∈ N
i.e. 2^3m – 1 is divisible by 7 ⇒ 2^3m – 1 = 7k; where k ∈ N …(1)
Now, 2^3(m+1) – 1 = 2^3m . 2³ – 1 = 8 (7k + 1) – 1 [using (1)]
= 56k + 7 = 7 (8k + 1) = 7k’, which is divisible by 7
where k’ = 8k + 1 ∈ N
∴ P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N

Question 4.
3²ⁿ – 1 is divisible by 8.
Solution:
Let P (n) be the statement: 3²ⁿ – 1 is divisible by 8.
For n = 1 ; 3^2×1 – 1 = 9 – 1 = 8, which is divisible by 8 ∴ P (1) is true.
Let us assume that P (m) is true, where m∈N
i.e. 3^2m – 1 is divisible by 8 ∴ 3^2m – 1 = 8k where k ∈ N …(1)
Now 3^2(m+1) – 1 = 3^2m.3² – 1 = 9 (8k + 1) – 1 [using (1)]
= 72k + 8 = 8 (9k + 1) = 8k’, where 9k + 1 = k’ ∈ N
which is divisible by 8
∴ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

Question 5.
4ⁿ + 15n – 1 is divisible by 9.
Solution:
Let P (n) be the statement: 4ⁿ + 15n – 1 is divisible by 9
For n = 1 ; 4¹ + 15.1 – 1 = 18, which is divisible by 9 ∴ P (1) is true.
Let us assume that P (m) is true where m ∈ N
i.e. 4^m + 15m – 1 is divisible by 9
∴ 4^m + 15m – 1 = 9k, where k ∈ N …(1)
Now we shall prove that P (m + 1) is true.
Now, 4^m+1 + 15 (m + 1) – 1 = 4^m . 4 + 15 (m + 1) – 1
= 4 (9k – 15m + 1) + 15m + 15 – 1 [using eqn. (1)]
= 36k – 60m + 4 + 15m + 14 = 36k – 45m +18 = 9 (4k – 5m + 2) = 9k’
where k’ = 4k – 5m + 2 ∈ N
which is divisible by 9
∴ P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 6.
3⁴ⁿ⁺² + 5²ⁿ⁺¹ is a multiple of 14.
Solution:
Let P (n) be the statement: 3⁴ⁿ⁺² + 5²ⁿ⁺¹ is a multiple of 14.
For n = 1 ; 3^4.1+2 + 5^2.1+1 = 3⁶ + 5³ = 729 + 125 = 854
which is clearly multipleof 14. ∴ P (4) is true.
Let us assume that P (m) is true, where m∈N
i.e. 3^4m+2 + 5^2m+1 is a multiple of 14.
∴ 3^4m+2 + 5^2m+1 = 14k, where k ∈ N … (1)
Now we want to prove that P (m + 1) is true.
3^4(m+1)+2 + 5^2(m+1)+1 = 3^4m+2.3⁴ + 5^2m.5³ = (14k – 5^2m+1) 81 + 5^2m.5³
= 14k x 81 + 5^2m (125 – 405)
= 14k x 81 + 5^2m (- 280) = 14 [81k – 20 x 5^2m]
= 14k’
where k’ = 81k – 20 x 5^2m
which is divisible by 14
∴ P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 7.
7²ⁿ + (2^3n-3) 3^n-1 is divisible by 25.
Solution:
Let P (n) be the statement: 7²ⁿ + (2^3n-3) 3^n-1 is divisible by 25.
For n = 1 ; 7^2×1 + 2^3×1-3 31-1 =49 + 1 = 50
which is clearly divisible by 25 ∴ P (1) is true.
Let us assume that P (m) is true, where m ∈ N
i.e. 7^2m + (2^3m-3) 3^m-1 is divisible by 25
∴ 7^2m + 2^3m-3) . 3^m-1 = 25k; where k ∈ N … (1)
Now we shall prove that P (m + 1) is true.
Now, 7^2(m+1) + 2^3(m+1)-3) 3^m+1-1 = 7^2m.7² + 2^3m-3.2³.3^2m
= 7^2m.7² + 2^3m-3.3^m-1.24 = 24 [7^2m + 2^m-3.3^m-1] + 25.7^2m
= 24 x 25k + 25.7^2m [using eqn. (1)]
= 25 [24k + 7^2m], which is clearly divisible by 25 ∴ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

Question 8.
xⁿ – yⁿ is divisible by x – y when n is an even positive integer.
Solution:
Let P (n) be the statement: xⁿ – yⁿ is divisible byx-y
For n = 1 ; x’ – y’= x – y which is clearly divisible by x – y ∴ P (1) is true.
Let us assume that P (m) is true where m ∈ N.
i.e. x^m – y^m is divisible by (x – y)
∴ x^m – y^m = f(x, y)(x – y) …(1)
where f(x, y) be polynomial in x and y.
Now we shall prove that P (m + 1) is true.
Now, x^m+1 – y^m+1 = x^m. x – x^m. y + x^my – y^m.y = x^m(x – y) + y(x^m – y^m)
= x^m(x – y) + y f(x, y) (x – y) [using eqn. (1)]
= (x – y)[x^m + y f(x, y)]
which is clearly divisible by (x – y) ∴ P (m + 1) is true.
Hence by mathematical induction P (m) is true for all n ∈ N.

Question 9.
Prove by the method of mathematical induction that 3^2n-1 + 3ⁿ + 4 is divisible by 2 for all n ∈ N.
Solution:
Let P (n) be the statement: 3^2n-1 + 3ⁿ + 4 is divisible by 2
For n = 1 ; 3^2×1-1 + 3¹ + 4 = 3 + 3 + 4 = 10, which is divisible by 2
∴ P(1) is true.
Let us assume that P (m) is true where m ∈ N
i.e. 3^2m-1 + 3^m + 4 is divisible by 2 ⇒ 3^m-1 + 3^m + 4 = 2k, where k ∈ N …(1)
Now we shall prove that P (m + 1) is true.
3^2(m+1)-1+ 3^m+1 + 4 = 3^2m-1 . 3² + 3^m . 3 + 4 = 3² [2k – 3^m – 4] + 3^m.3 + 4
= 18k – 3^m(9 – 3) – 36 + 4 = 18k – 6.3^m – 32
= 2 [9k – 3.3^m – 16]
which is divisible by 2
∴ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.
Prove by the principle of mathematical induction.

Question 10.
n < 2ⁿ for all n ∈ N
Solution:
Let P (n) be the statement : n < 2ⁿ
For n = 1 ; 1 < 2¹ ∴ P(1) is true.
Let us assume that P (m) is true, where m ∈ N
i.e. m < 2ⁿ … (1)
Now we shall prove that P (m + 1) is true.
P (m + 1) : m + 1 < 2^m + 1
< 2^m + 2^m = 2.2^m = 2^m+1 [∵ 1 < 2^m ∀ m ∈ N]
Hence by mathematical induction, P (n) is true for all n ∈ N

Question 11.
(2n + 7) < (n + 3)² for all natural numbers.
Solution:
Let P (n) be the statement: (2n + 7) < (n + 3)²
For n = 1 ; 2 x 1 + 7 = 9 < 16 = (1 + 3)² which is true
∴ P(1) is true.
Let us assume that P (m) is true, where m ∈ N
i.e. (2m + 7) < (m + 3)² … (1)
Now, 2 (m + 1) + 7 = 2m + 7 + 2 < (m + 3)² + 2 [using (1)]
= m² + 6m + 9 + 2 = m² + 8m + 16 – (2m + 5) = (m + 4)² – (2m + 5)
⇒ 2 (m + 1) + 7 < (m + 4)² – (2m + 5) < (m + 4)² ∴ P (w + 1) is true. Hence by mathematical induction, P (n) is true for all n ∈ N Question 12. 1² + 2² + 3² + … + n² > \(\frac{n^3}{3}\) for all n ∈ N.
Solution:
Let P (n) be the statement: 1² + 2² + 3² + … + n² > \(\frac{n^3}{3}\)
For n = 1 ; 1² 1 > \(\frac { 1 }{ 3 }\) = \(\frac { 1³ }{ 3 }\) ∴ P(1) is true.
Let us assume that P (m) is true where m ∈ N

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