Questions & Answers

Question

Answers

A. $\dfrac{5n-4}{5n+6}$

B. $\dfrac{n}{5n+6}$

C. $\dfrac{2n-1}{5n+6}$

D. $\dfrac{1}{5n+6}$

Answer
Verified

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 $\dfrac{n}{5n+6}$.