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Prove the following by using the principle of mathematical induction for all $n \in N$ \[\] $ \large\frac{1}{2.5}$$+\large\frac{1}{5.8}$$+\large\frac{1}{8.11}$$+...+\large\frac{1}{(3n-1)(3n+2)}$$=\large\frac{1}{(6n+4)}$

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Let the given statement be $P(n)$, i.e.,
$P(n) : \large\frac{1}{2.5}$$+\large\frac{1}{5.8}$$+\large\frac{1}{8.11}$$+...+\large\frac{1}{(3n-1)(3n+2)}$$=\large\frac{1}{(6n+4)}$
For $n = 1$, we have
$P(1) = \large\frac{1}{2.5}$$=\large\frac{1}{10}$$=\large\frac{1}{6.1+4}$$=\large\frac{1}{10}$, which is true.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$ \large\frac{1}{2.5}$$+\large\frac{1}{5.8}$$+\large\frac{1}{8.11}$$+...+\large\frac{1}{(3k-1)(3k+2)}$$=\large\frac{1}{(6k+4)}$-------------(i)
We shall now prove that $P(k+1)$ is true.
Consider
$= \large\frac{1}{2.5}$$+\large\frac{1}{5.8}$$+\large\frac{1}{8.11}$$+...+\large\frac{1}{(3k-1)(3k+2)}$$+\large\frac{1}{\{ 3(k+1)-1\} \{3 (k+1)+2 \}}$
$= \large\frac{k}{6k+4}$$+\large\frac{1}{(3k+3-1)(3k+3+2)}\qquad$ [ Using (i) ]
$= \large\frac{k}{6k+4}$$+\large\frac{1}{(3k+2)(3k+5)}$
$= \large\frac{k}{2(3k+2)}$$+\large\frac{1}{(3k+2)(3k+5)}$
$= \large\frac{1}{(3k+2)}$$\bigg( \large\frac{k}{2}$$+\large\frac{1}{3k+5} \bigg)$
$= \large\frac{1}{(3k+2)}$$\bigg( \large\frac{k(3k+5)+2}{2(3k+5)} \bigg)$
$= \large\frac{1}{(3k+2)}$$\bigg( \large\frac{3k^2+5k+2}{2(3k+5)} \bigg)$
$= \large\frac{1}{(3k+2)}$$\bigg( \large\frac{(3k+2)(k+1)}{2(3k+5)} \bigg)$
$ = \large\frac{k+1}{6k+10}$
$ = \large\frac{k+1}{6(k+1)+4}$
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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