Question

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

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

Answer 1

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

We have $t_n=\frac{n+2}{n(n+1)}.{\left(\frac{1}{2}\right)}^n=\frac{2(n+1)-n}{n(n+1)}{\left(\frac{1}{2}\right)}^n$

$=\frac{1}{n}{\left(\frac{1}{2}\right)}^{n-1}-\frac{1}{n-1}{\left(\frac{1}{2}\right)}^n$

$\therefore S_n=\sum _{n=1}^n{t}_n=\{\frac{1}{1}\left(\frac{1}{2}\right)-\frac{1}{2}{\left(\frac{1}{2}\right)}^1\}+\{\frac{1}{2}{\left(\frac{1}{2}\right)}^1-\frac{1}{3}{\left(\frac{1}{2}\right)}^2\}$

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