Question

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

• A. $\frac{n}{6n+4}$
• B. $\frac{n}{3n+2}$
• C. $\frac{n}{4n+6}$
• D. $\frac{1}{2(2n+3)}$

Answer 1

Answer: (A)

$\frac{n}{6n+4}$

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

After expanding terms will get canceled out and remaining are
$=\frac{1}{3.2}-\frac{1}{3(3n+2)}$

$=\frac{n}{6n+4}$

Hence, option 'A' is correct.