Question

The sum of the series
$\frac{2}{1\times 2}+\frac{5}{2\times 3}\times 2+\frac{10}{3\times 4}\times 2^2+\frac{17}{4\times 5}\times 2^3+----$
$--$ upto $9$ terms is given as

• A. $0.99\times 2^{10}$
• B. $0.9\times 2^9$
• C. $1.1\times 2^{10}$
• D. $1.1\times 2^9$

Answer 1

The correct option is B $0.9\times 2^9$

The ${}^{\prime }{r}^{\prime }$th term of the series can be written as
$T_r=\frac{r^2+1}{r(r+1)}\times 2^{r-1}=\frac{r^2+1+2r-2r}{r(r+1)}\times 2^{r-1}=[\frac{r+1}{r}-\frac{2}{r+1}]\times 2^{r-1}=2^{r-1}+[\frac{1}{r}-\frac{2^r}{r+1}]\times 2^{r-1}=2^{r-1}+[\frac{2^{r-1}}{r}-\frac{2^r}{r+1}]\text{Let}\frac{2^{r-1}}{r}=t_r\Rightarrow T_r=2^{r-1}+[t_r-t_{r+1}]$
Sum of $T_r$ upto $n$ terms -
$S_n=\sum _{r=1}^n{2}^{r-1}+[t_1-t_2+t_2-t_3+t_3+---+t_{n-1}-t_n]=1\times \frac{2^n-1}{2-1}+[\frac{2^0}{1}-\frac{2^n}{n+1}]=2^n\left(\frac{n}{n+1}\right)\Rightarrow S_9=0.9\times 2^9$