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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}$

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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.
$\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 by thanvigandhi_1

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