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Prove the following by using the principle of mathematical induction for all $n \in N$ \[\] $ \large\frac{1}{3.5}$$+\large\frac{1}{5.7}$$+\large\frac{1}{7.9}$$+...+\large\frac{1}{(2n+1)(2n+3)}$$=\large\frac{n}{3(2n+3)}$

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Let the given statement be $P(n)$, i.e.,
$P(n) : \large\frac{1}{3.5}$$+\large\frac{1}{5.7}$$+\large\frac{1}{7.9}$$+...+\large\frac{1}{(2n+1)(2n+3)}$$=\large\frac{n}{3(2n+3)}$
For $n = 1$, we have
$ P(1) : \large\frac{1}{3.5}$$=\large\frac{1}{3(2.1+3)}$$=\large\frac{1}{3.5}$, which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$P(k) : \large\frac{1}{3.5}$$+\large\frac{1}{5.7}$$+\large\frac{1}{7.9}$$+...+\large\frac{1}{(2k+1)(2k+3)}$$=\large\frac{n}{3(2k+3)}$---------(1)
We shall now prove that $P(k+1)$ is true.
Consider
$\bigg[ \large\frac{1}{3.5}$$+\large\frac{1}{5.7}$$+\large\frac{1}{7.9}$$+...+\large\frac{1}{(2k+1)(2k+3)}\bigg]$$+ \large\frac{1}{\{2(k+1)+1\}\{2(k+1)+3\}}$
$ = \large\frac{k}{3(2k+3)}$$+ \large\frac{1}{(2k+3)(2k+5)}\qquad$ [ Using (1)]
$= \large\frac{1}{(2k+3)}$$ \bigg[ \large\frac{k}{3}$$+\large\frac{1}{(2k+5)} \bigg]$
$= \large\frac{1}{(2k+3)}$$ \bigg[ \large\frac{k(2k+5)+3}{3(2k+5)} \bigg]$
$= \large\frac{1}{(2k+3)}$$ \bigg[ \large\frac{2k^2+5k+3}{3(2k+5)} \bigg]$
$= \large\frac{1}{(2k+3)}$$ \bigg[ \large\frac{2k^2+2k+3k+3}{3(2k+5)} \bigg]$
$= \large\frac{1}{(2k+3)}$$ \bigg[ \large\frac{2k(k+1)+3(k+1)}{3(2k+5)} \bigg]$
$ = \large\frac{(k+1)(2k+3)}{3(2k+3)(2k+5)}$
$= \large\frac{(k+1)}{3\{ 2(k+1)+3\}}$
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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