Question

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..n$ terms $=$

• A. $\frac{3n}{2(3n+2)}$
• B. $\frac{3n}{3n+2}$
• C. $\frac{n}{2(3n+2)}$
• D. $\frac{n}{3n+2}$

Answer 1

Answer: (C)

$\frac{n}{2(3n+2)}$

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.....+\frac{1}{[2+3(n-1)].[5+3(n-1)]}$

$={\sum }_{n=1}^n\frac{1}{[2+3n-3].[5+3n-3]}$

$={\sum }_{n=1}^n\frac{1}{(3n-1).(3n+2)}=\frac{1}{3}\sum \frac{[3n+2-(3n-1)]}{(3n-1)(3n+2)}$

$=\frac{1}{3}\sum \frac{1}{3n-1}-\frac{1}{3n+2}$

$=\frac{1}{3}[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3n-4}-\frac{1}{3n-1}+\frac{1}{3n-1}-\frac{1}{3n+2}]$

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