For, n = 1

LHS of P (n) = 1

RHS of P (n) = \(\dfrac{ 1 (1+1)}{2}\) = 1

So, LHS = RHS

Since, P (n) is true for n = 1

Let us consider P (n) be the true for n = k, so

1 + 2 + 3 + …. + k = \(\dfrac{ k (k+1)}{2}\) … (i)

Now,

(1 + 2 + 3 + … + k) + (k + 1)

= \( \dfrac{ k (k+1)}{2}\) + (k+1)

= (k + 1) (\( \dfrac{k}{2}\) + 1)

=\(\dfrac{ [(k+1) [(k+1) + 1]] }{ 2}\)

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

P (n) is true for all n ∈ N

So, by the principle of Mathematical Induction

Hence, P (n) = 1 + 2 + 3 + ….. + n = \(\dfrac{ n (n +1)}{2}\) is true for all n ∈ N.

Answered by Aaryan | 2 years 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.

(ab)^{ n} = a^{n} b^{n} for all n ϵ N