Question

The sum of $\frac{3}{1.2}.\left(\frac{1}{2}\right)+\frac{4}{2.3}.{\left(\frac{1}{2}\right)}^2+\frac{5}{3.4}.{\left(\frac{1}{2}\right)}^3+...$ to n terms is equal to

• A. $1-\frac{1}{(n+1)2^n}$
• B. $1-\frac{1}{n{2}^{n-1}}$
• C. $1+\frac{1}{(n+1)2^n}$
• D. none of these

Answer 1

The correct option is A $1-\frac{1}{(n+1)2^n}$

Given that, $S_n=\frac{3}{1.2}.\left(\frac{1}{2}\right)+\frac{4}{2.3}.{\left(\frac{1}{2}\right)}^2+\frac{5}{3.4}.{\left(\frac{1}{2}\right)}^3+...+\left(\frac{n+2}{n(n+1)}\right){\left(\frac{1}{2}\right)}^n$
Now, $t_r=\left(\frac{r+2}{r(r+1)}\right){\left(\frac{1}{2}\right)}^r=2[\frac{1}{r{2}^r}-\frac{1}{(r+1)2^{r+1}}]$
Therefore, $S_n={\sum }_{r=1}^n{t}_r=2{\sum }_{r=1}^n[\frac{1}{r{2}^r}-\frac{1}{(r+1)2^{r+1}}]$
$\Rightarrow S_n=2[\frac{1}{2}-\frac{1}{2.2^2}+\frac{1}{2.2^2}-\frac{1}{3.2^3}+......+\frac{1}{{n2}^n}-\frac{1}{(n+1)2^{n+1}}]$
$\Rightarrow S_n=2[\frac{1}{2}-\frac{1}{(n+1)2^{n+1}}]=1-\frac{1}{(n+1)2^n}$
Ans: A