Let us check for n = 1,

P (1): \(\dfrac{ 1}{2^1} = 1 – \dfrac{1}{2^1}\)

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

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.

Let P (k): \(\dfrac{1}{2}+ \dfrac{1}{4}+ \dfrac{1}{8} + … + \dfrac{1}{2^k} = 1 – \dfrac{1}{2^k}\) … (i)

So,

\( \dfrac{1}{2}+ \dfrac{1}{4}+ \dfrac{1}{8} + … + \dfrac{1}{2^k} +\dfrac{1}{2^{k+1}}\)

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

= \( 1 – \dfrac{1}{2^k} + \dfrac{1}{2k+1}\) by using equation (i)

= \( 1 – \dfrac{(2-1)}{2^{(k+1)}}\)

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.

(ab)^{ n} = a^{n} b^{n} for all n ϵ N