Let the given statement be P(n), i.e., P (n) : $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$ + ... + $\frac{1}{2^n}$ = 1 - $\frac{1}{2^n}$ First we prove that the statement is true for n = 1 P (1) : $\frac{1}{2}$ = 1 - $\frac{1}{2}$ , which is true Assume P(k) is true for some positive integer k, i.e., $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$ + ... + $\frac{1}{2^k}$ = 1 - $\frac{1}{2^k}$ ... (i) We shall now prove that P(k + 1) is also true. For this we have to prove that $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$ + ... + $\frac{1}{2^k} + \frac{1}{2^k + 1}$ = 1 - $\frac{1}{2^k+1}$ L.H.S. = $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$ + ... + $\frac{1}{2^k} + \frac{1}{2^k + 1}$ = 1 - $\frac{1}{2^k} - \frac{1}{2^k + 1}$ [From (i)] = 1 - $\left[\frac{1}{2^k} - \frac{1}{2^k+1}\right]$ = 1 - $\frac{1}{2^k}\left[1-\frac{1}{2}\right]$ = 1 - $\frac{1}{2^k} \cdot \frac{1}{2}$ = 1 - $\frac{1}{2^k+1}$ = R.H.S. Thus P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true ∀ n ∈ N.