Let P (n): (ab)^{ n} = a^{n} b^{n}

Let us check for n = 1,

P (1): (ab)^{ 1} = a^{1} b^{1}

: ab = ab

P (n) is true for n = 1.

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): (ab)^{ k} = a^{k} b^{k} … (i)

We have to prove,

(ab) ^{k + 1 }= a^{k + 1}.b^{k + 1}

So,

= (ab)^{ k + 1}

= (ab)^{ k} (ab)

= (a^{k }b^{k}) (ab) using equation (1)

= (a^{k + 1}) (b^{k + 1})

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

Hence, P (n) is true for all n ∈ N.

Answered by Aaryan | 1 year agoGiven an example of a statement P (n) such that it is true for all n ϵ N.

a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]

7^{2n} + 2^{3n – 3}. 3n – 1 is divisible by 25 for all n ϵ N

n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.

3^{2n + 2} – 8n – 9 is divisible by 8 for all n ϵ N