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Prove the following by using the principle of mathematical induction for all $n \in N$ : \[\] $1+3+3^2+...+3^{n-1}= \large\frac{(3^n-1)}{2}$

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Let the given statement be $P(n)$, i.e.,
For $n=1$ we have,
$P(1):1 = \large\frac{(3^1-1)}{2}$$=\large\frac{3-1}{2}$$=\large\frac{2}{2}$$=1$, which is true
Let $P(k)$ be true for some positive integer $k$, i.e.,
$1+3+3^2+...+3^{k-1}= \large\frac{(3^k-1)}{2}$-----------(i)
We shall now prove that $P(k+1)$ is true.
Consider
$1+3+3^2+...+3^{k-1}+3^{(k+1)-1}$
$(1+3+3^2+...+3^{k-1})+3^k$
$ = \large\frac{(3^k-1)}{2}$$ +3^k \qquad $ [ Using (i) ]
$ = \large\frac{(3^k-1)+2.3^k}{2}$
$ = \large\frac{(1+2)3^k-1}{2}$
$ = \large\frac{3.3^k-1}{2}$
$ = \large\frac{3^{k+1}-1}{2}$
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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