# Prove the following by using the principle of mathematical induction for all $n \in N$  $\large\frac{1}{2}$$+ \large\frac{1}{4}$$+\large\frac{1}{8}$$+...+\large\frac{1}{2^n}$$=1-\large\frac{1}{2^n}$

Let the given statement be $P(n)$, i.e.,
$P(n) : \large\frac{1}{2}$$+ \large\frac{1}{4}$$+\large\frac{1}{8}$$+...+\large\frac{1}{2^n}$$=1-\large\frac{1}{2^n}$
For $n = 1$, we have
$P(1) : \large\frac{1}{2} $$= 1-\large\frac{1}{2^1}$$= \large\frac{1}{2}$, which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$\large\frac{1}{2}$$+ \large\frac{1}{4}$$+\large\frac{1}{8}$$+...+\large\frac{1}{2^k}$$=1-\large\frac{1}{2^k}$-----------(i)
We shall now prove that $P(k+1)$ is true.
Consider
$\bigg( \large\frac{1}{2}$$+ \large\frac{1}{4}$$+\large\frac{1}{8}$$+...+\large\frac{1}{2^k}\bigg)$$+\large\frac{1}{2^{k+1}}$
$\bigg( 1 - \large\frac{1}{2^k} \bigg)$$+\large\frac{1}{2^{k+1}}$$\qquad$ [ Using (i) ]
$= 1-\large\frac{1}{2^k}+\large\frac{1}{2.2^k}$
$= 1-\large\frac{1}{2^k}$$\bigg( 1- \large\frac{1}{2} \bigg) = 1-\large\frac{1}{2^k}$$\bigg( \large\frac{1}{2} \bigg)$
$= 1-\large\frac{1}{2^{k+1}}$
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.
answered Apr 29, 2014