Question

If $S_n=\frac{1}{6\times 11}+\frac{1}{11\times 16}+\frac{1}{16\times 21}.............$up to $n$ terms, then $6S_n$ equals

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

Answer 1

The correct option is B

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

Explanation for the correct option

Step 1: We can write,

$\frac{1}{6\times 11}=\frac{1}{5}\left(\frac{1}{6}-\frac{1}{11}\right)$ …$\left(1\right)$

Similarly,

$\frac{1}{11\times 16}=\frac{1}{5}\left(\frac{1}{11}-\frac{1}{16}\right)...\left(2\right)$

$\frac{1}{16\times 21}=\frac{1}{5}\left(\frac{1}{16}-\frac{1}{21}\right)$ ….$\left(3\right)$

And so on.

Step 2: Let $S_n=\frac{1}{6\times 11}+\frac{1}{11\times 16}+\frac{1}{16\times 21}.............n$ terms

Then,

$S_n=\frac{1}{6\times 11}+\frac{1}{11\times 16}+\frac{1}{16\times 21}.............\frac{1}{\left(5n+1\right)\left(5n+6\right)}$

Step 3: From equation $\left(1\right)\mathbf{,}\mathbf{}\left(2\right)\mathbf{}\mathbf{\text{and}}\mathbf{}\left(3\right)$ we can write $S_n$,

$S_n=\frac{1}{5}\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}..............-\frac{1}{5n+6}\right)$

$S_n=\frac{1}{5}\left(\frac{1}{6}-\frac{1}{5n+6}\right)$

$S_n=\frac{1}{5}\left(\frac{5n+6-6}{6\left(5n+6\right)}\right)$

$S_n=\left(\frac{n}{6\left(5n+6\right)}\right)$

$6S_n=\frac{n}{5n+6}$

Hence, the correct option is ‘B’.