# 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)}$

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)$