Question

The sum of $\frac{3}{1\cdot 2}\left(\frac{1}{2}\right)+\frac{4}{2\cdot 3}{\left(\frac{1}{2}\right)}^2+\frac{5}{3\cdot 4}{\left(\frac{1}{2}\right)}^3+\cdots$ upto $n$ terms is

• A. $1-\frac{1}{n+1}{\left(\frac{1}{2}\right)}^n$
• B. $1+\frac{1}{n+1}{\left(\frac{1}{2}\right)}^n$
• C. $1-\frac{1}{n+1}{\left(\frac{1}{2}\right)}^{n-1}$
• D. $1+\frac{1}{n+1}{\left(\frac{1}{2}\right)}^{n-1}$

Answer 1

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

$S_n=\frac{3}{1\cdot 2}\left(\frac{1}{2}\right)+\frac{4}{2\cdot 3}{\left(\frac{1}{2}\right)}^2+\frac{5}{3\cdot 4}{\left(\frac{1}{2}\right)}^3+\cdots$
General term of the series is
$T_n=\frac{n+2}{n(n+1)}{\left(\frac{1}{2}\right)}^n=\frac{\left[2\right(n+1)-n]}{n(n+1)}{\left(\frac{1}{2}\right)}^n=\frac{1}{n}{\left(\frac{1}{2}\right)}^{n-1}-\frac{1}{n+1}{\left(\frac{1}{2}\right)}^n$
Which is the form of $V_n-V_{n+1}$
Hence
$S_n=V_1-V_{n+1}=1-\frac{1}{n+1}{\left(\frac{1}{2}\right)}^n$

Alternate Solution:
put the values of $n$ as $1,2,3,\cdots$ in the options
and check with the series given in the question.