Question

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

Answer 1

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

The $n^{th}$ term of the series, $t_n$, can be written as
$t_n=\frac{n+2}{n(n+1)}{\left(\frac{1}{2}\right)}^n$
$\Rightarrow t_n=\frac{2(n+1)-n}{n(n+1)}{\left(\frac{1}{2}\right)}^n$
$\Rightarrow t_n=\left(\frac{2}{n}-\frac{1}{n+1}\right){\left(\frac{1}{2}\right)}^n$
$\therefore t_1=\left(\frac{2}{1}-\frac{1}{2}\right){\left(\frac{1}{2}\right)}^1=1-\frac{1}{4}$
$t_2=(\frac{2}{2}-\frac{1}{3}){\left(\frac{1}{2}\right)}^2=\frac{1}{4}-\frac{1}{12}$
$t_3=(\frac{2}{3}-\frac{1}{4}){\left(\frac{1}{2}\right)}^3=\frac{1}{12}-\frac{1}{32}$
$t_4=(\frac{2}{4}-\frac{1}{5}){\left(\frac{1}{2}\right)}^4=\frac{1}{32}-\frac{1}{80}$
.
$t_n=(\frac{2}{n}-\frac{1}{n+1}){\left(\frac{1}{2}\right)}^n$

On adding all terms, we get
$S_n=1-\frac{1}{(n+1)2^n}$
Hence, option A is correct.