Question

The sum of $\frac{7}{2\times 3}\left(\frac{1}{3}\right)+\frac{9}{3\times 4}{\left(\frac{1}{3}\right)}^2+\frac{11}{4\times 5}{\left(\frac{1}{3}\right)}^3+\dots$ upto $10$ terms is equal to

• A. $\frac{1}{2}-\frac{1}{12\times 3^{10}}$
• B. $\frac{1}{3}-\frac{1}{12\times 3^{10}}$
• C. $\frac{1}{2}-\frac{1}{10\times 3^{10}}$
• D. none of these

Answer 1

The correct option is A $\frac{1}{2}-\frac{1}{12\times 3^{10}}$

$\frac{7}{2\times 3}\left(\frac{1}{3}\right)+\frac{9}{3\times 4}{\left(\frac{1}{3}\right)}^2+\frac{11}{4\times 5}{\left(\frac{1}{3}\right)}^3+....$

General term $T_n=\frac{2n+5}{(n+1)(n+2)}{\left(\frac{1}{3}\right)}^n$

It can be simplified as

$T_n=\left[\frac{3(n+2)-(n+1)}{(n+1)(n+2)}\right]{\left(\frac{1}{3}\right)}^n$

$=[\frac{3}{n+1}-\frac{1}{n+2}]{\left(\frac{1}{3}\right)}^n$

Sum to n terms

$S_{10}=\sum _{n=1}^{10}{T}_n=\sum _{n=1}^{10}(\frac{3}{n+1}-\frac{1}{n+1}){\left(\frac{1}{3}\right)}^n$

$=(\frac{3}{2}-\frac{1}{3})\left(\frac{1}{3}\right)+(\frac{3}{3}-\frac{1}{4}){\left(\frac{1}{3}\right)}^2+(\frac{3}{4}-\frac{1}{5}){\left(\frac{1}{3}\right)}^3+......+(\frac{3}{11}-\frac{1}{12}){\left(\frac{1}{3}\right)}^{10}$

$=\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{36}+\frac{1}{36}......+\frac{3}{11}\times \frac{1}{3^{10}}-\left(\frac{1}{12}\right){\left(\frac{1}{3}\right)}^{10}$

$s_{10}=\frac{1}{2}-\frac{1}{12}{\left(\frac{1}{3}\right)}^{10}$.