Question

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

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

Answer 1

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

$t_n=\frac{n+2}{n(n+1)}\cdot {\left(\frac{1}{2}\right)}^n$

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

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

$S_n={\sum }_{n=1}^n{t}_n$

$=\{\frac{1}{1}{\left(\frac{1}{2}\right)}^0-\frac{1}{2}{\left(\frac{1}{2}\right)}^1\}$ $+\{\frac{1}{2}{\left(\frac{1}{2}\right)}^1-\frac{1}{3}{\left(\frac{1}{2}\right)}^2\}+...+$ $=\{\frac{1}{n}{\left(\frac{1}{2}\right)}^{n-1}-\frac{1}{n+1}{\left(\frac{1}{2}\right)}^n\}$

$S_n=1-\frac{1}{(n+1)2^n}$