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

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Let the given statement be $P(n)$, i.e.,
$P(n) : a+ar+ar^2+...+ar^{n-1}=\large\frac{a(r^n-1)}{r-1}$
For $n = 1$, we have
$P(1) : a=\large\frac{a(r^1-1)}{r-1}$$=a$, which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$ a+ar+ar^2+...+ar^{k-1}=\large\frac{a(r^k-1)}{r-1}$--------(i)
We shall now prove that $P(k+1)$ is true.
Consider
$\{ a+ar+ar^2+...+ar^{k-1} \}+ar^{(k+1)-1}$
$ = \large\frac{a(r^k-1)}{r-1}$$+ar^k \qquad$ [ Using (i) ]
$ = \large\frac{a(r^k-1)+ar^k(r-1)}{r-1}$
$ = \large\frac{a(r^k-1)+ar^{k+1}-ar^k}{r-1}$
$ = \large\frac{ar^k-a+ar^{k+1}-ar^k}{r-1}$
$= \large\frac{ar^{k+1}-a}{r-1}$
$ = \large\frac{a(r^{k+1}-1)}{r-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 30, 2014 by thanvigandhi_1
 

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