Question
Answers

If ${{S}_{n}}=\dfrac{1}{6.11}+\dfrac{1}{11.16}+\dfrac{1}{16.21}+..........$ to $n$ terms, then $6{{S}_{n}}$ is equal to
A. $\dfrac{5n-4}{5n+6}$
B. $\dfrac{n}{5n+6}$
C. $\dfrac{2n-1}{5n+6}$
D. $\dfrac{1}{5n+6}$

Answer Verified Verified
Hint: In order to solve this type of question, we have to get the pattern of rth term i.e. ${{T}_{r}}$. Carefully looking the pattern we can say that rth term is of the form, ${{T}_{r}}=\dfrac{1}{\left( 5r+1 \right)\left( 5r+6 \right)}$. After getting the rth term pattern, sum will be equal to it’s summation i.e. ${{S}_{n}}=\sum{{{T}_{r}}}$. In order to find the summation, break ${{T}_{r}}$ into partial fractions and then when we sum it then adjacent terms will cancel out only the first and last term will remain. So after doing that find ${{S}_{n}}$ and then multiply it with 6 and that will be our answer.

Complete step-by-step answer:
Here it is given ${{S}_{n}}=\dfrac{1}{6.11}+\dfrac{1}{11.16}+\dfrac{1}{16.21}+..........$ to $n$ terms. In order to find ${{S}_{n}}$, first we have to find the pattern of rth term.
Let’s observe the pattern
$6=5\times 1+1$ and $11=5\times 1+6$
$11=5\times 2+1$ and $16=5\times 2+6$
$16=5\times 3+1$ and $21=5\times 3+6$
Hence, with this we can say that ${{T}_{r}}=\dfrac{1}{\left( 5r+1 \right)\left( 5r+6 \right)}$
Splitting ${{T}_{r}}$ in partial fractions, we get
$\Rightarrow {{T}_{r}}=\dfrac{1}{5}\left( \dfrac{5}{\left( 5r+1 \right)\left( 5r+6 \right)} \right)=\dfrac{1}{5}\left( \dfrac{1}{\left( 5r+1 \right)}-\dfrac{1}{\left( 5r+6 \right)} \right)$
Now, we know that ${{S}_{n}}=\sum{{{T}_{r}}}$.
\[\Rightarrow {{S}_{n}}=\sum\limits_{r=1}^{r=n}{\dfrac{1}{5}\left( \dfrac{1}{\left( 5r+1 \right)}-\dfrac{1}{\left( 5r+6 \right)} \right)}\]
Expanding the summation, we get
$\Rightarrow {{S}_{n}}=\dfrac{1}{5}\left( \dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{21}+\dfrac{1}{21}+.............-\dfrac{1}{5n+1}+\dfrac{1}{5n+1}-\dfrac{1}{5n+6} \right)$
It is clear that adjacent term will cancel out in the above expression, only the two terms last and first will remain, so
$\Rightarrow {{S}_{n}}=\dfrac{1}{5}\left( \dfrac{1}{6}-\dfrac{1}{5n+6} \right)$
$\Rightarrow {{S}_{n}}=\dfrac{1}{5}\left( \dfrac{5n+6-6}{6(5n+6)} \right)$
$\Rightarrow {{S}_{n}}=\dfrac{n}{6(5n+6)}$
Hence, we get the value of ${{S}_{n}}=\dfrac{n}{6(5n+6)}$.
Now, we have to calculate the value of $6{{S}_{n}}$ which is equal to $\dfrac{n}{5n+6}$.

So, the correct answer is “Option (B)”.

Note: This is a typical question of progression and series where the main crux is to notice two things, first one is finding the pattern and the second one is to notice the difference between the denominators are constant i.e. $\left( 11-6 \right)=\left( 16-11 \right)=\left( 21-16 \right)=..........=5$. Students should notice these two things and apply the standard process that is getting ${{T}_{r}}$ and break it into partial fractions in order to get the sum and get the correct answer.