Question

Prove that $\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{n}{(3n+1)}$

Answer 1

For $1st$ part of denominator,

$1,4,7,11\dots a=1,d=3$

$a_n=a+(n-1)d$

$a_n=1+(n-1)3$

$a_n=(3n-2)$

For $2nd$ part of denominator,

$4,7,10\dots$

$a=4,d=3$

$a_n=4+(n-1)3$

$a_n=4+3n-3=(3n+1)$

$a_n=\frac{1}{(3n-2)(3n+1)}$

$a_n=\frac{1}{3}\left[\frac{3}{(3n-2)(3n+1)}\right]$

$a_n=\frac{1}{3}\left[\frac{(3n+1)-(3n-2)}{(3n-2)(3n+1)}\right]$

$a_n=\frac{1}{3}[\frac{1}{3n-2}-\frac{1}{(3n+1)}]$

$a_n=\frac{1}{3}[1-\frac{1}{3n+1}]=\left(\frac{n}{3n+1}\right)$