Question

If $S_n=\frac{1}{6\cdot 11}+\frac{1}{11\cdot 16}+\frac{1}{16\cdot 21}+\cdots$ to $n$ terms, then $6S_n$ is equal to

• A. $\frac{5n-4}{5n+6}$
• B. $\frac{n}{(5n+6)}$
• C. $\frac{2n-1}{5n+6}$
• D. $\frac{1}{(5n+6)}$

Answer 1

Answer: (B)

$\frac{n}{(5n+6)}$

Given that $S_n=\frac{1}{6\cdot 11}+\frac{1}{11\cdot 16}+\frac{1}{16\cdot 21}+\cdots +n$ terms
$=\frac{1}{5}(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\cdots +\frac{1}{5n+1}-\frac{1}{5n+6})$
$=\frac{1}{5}(\frac{1}{6}-\frac{1}{5n+6})$
$=\frac{1}{5}\left(\frac{5n+6-6}{6(5n+6)}\right)$
$=\frac{n}{6(5n+6)}$
$\Rightarrow 6S_n=\frac{n}{5n+6}$