Question

Prove the following by the principle of mathematical induction: 

7²ⁿ + 2^{3n – 3} . 3n – 1 is divisible by 25 for all n ϵ N

Answer 1

Let P(n): 7²ⁿ + 2^{3n - 3}.3^{n - 1} is divisible by 25 

For n = 1 

= 7² + 2°.3°

= 49 + 1 

= 50 

Therefor it is divisible by 25 

So, P(n) is true for n = 1 

Now, P(n) is true For n = k, 

So, we have to show that 7²ⁿ + 2^{3n - 3}.3^{n - 1} is divisible by 25 

= 7^2k + 2^{3k - 3}.3^{k - 1} = 25λ - - - - - - - (1)

Now, P(n) is true For n = k + 1, 

So, we have to show that 7^{2k + 1} + 2^3k .3^k is divisible by 25 

= 7^{2k + 2} + 2^3k .3^k = 25μ 

Now, 

= 7^{2(k + 1)} + 2^3k .3^k 

= 7^2k .7¹ + 2^3k .3^k 

= (25λ – 2^{3k - 3}.3^{k - 1})49 + 2^3k .3^k from eq 1

= 25μ 

Therefore, P(n) is true for n = k + 1 

Hence, P(n) is true for all n ∈ N