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

For n=1

= 7² + 2⁰.3⁰

= 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.3k from eq 1

=

= 24×25×49λ - 2^3k.3^k..49 + 24.2^3k.3^k

= 24×25×49λ - 25.2^3k.3^k

= 25(24.49λ - 2^3k.3^k)

= 25μ

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

Hence, P(n) is true for all n ∈ N