1/2 + 1/4 + 1/8 + … + 1/2n = 1 – 1/2n

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.