Let P(n): 7^{2n} + 2^{3n - 3}.3^{n - 1} is divisible by 25

For n = 1

= 7^{2} + 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 7^{2n} + 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} .7^{1} + 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