For, n = 1

P (n) = \( \dfrac{1}{1.2} = \dfrac{1}{1+1}\)

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

P (n) is true for n = 1

Let’s check for P (n) is true for n = k,

\(\dfrac{ 1}{1.2} + \dfrac{1}{2.3} +\dfrac{ 1}{3.4 }+ … + \dfrac{1}{k(k+1)} + \dfrac{k}{(k+1) (k+2)} = \dfrac{(k+1)}{(k+2)}\)

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

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

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.

(ab)^{ n} = a^{n} b^{n} for all n ϵ N