Let us check for n = 1,

P (1): a =\(\dfrac{ a (r^1 – 1)}{(r-1)}\)

a = a

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): a + ar + ar^{2} + … + ar^{k – 1} = \(\dfrac{ a (r^k – 1)}{(r – 1)}\) … (i)

So,

a + ar + ar^{2} + … + ar^{k – 1} + ar^{k}

Now, substituting the value of P (k) we get,

= \(\dfrac{a(rk – 1)}{(r – 1)}\) + ar^{k} by using equation (i)

=\( \dfrac{ a(rk+1 – 1)}{ (r-1)}\)

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

Hence, P (n) 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