Question

$\frac{\frac{1}{2}\cdot \frac{2}{2}}{1^3}+\frac{\frac{2}{2}\cdot \frac{3}{2}}{1^3+2^3}+\frac{\frac{3}{2}\cdot \frac{4}{2}}{1^3+2^3+3^3}+\cdots$ upto $n$ terms

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

Answer 1

The correct option is B $\frac{n}{n+1}$

$n^{th}$ term of the series $\frac{\frac{1}{2}\cdot \frac{2}{2}}{1^3}+\frac{\frac{2}{2}\cdot \frac{3}{2}}{1^3+2^3}+\frac{\frac{3}{2}\cdot \frac{4}{2}}{1^3+2^3+3^3}+\cdots$ is

$T_n=\frac{\frac{n(n+1)}{2\cdot 2}}{\sum n^3}=\frac{\frac{n(n+1)}{4}}{\frac{n^2{(n+1)}^2}{4}}$
$=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$
$\therefore S_n=\sum (\frac{1}{n}-\frac{1}{n+1})$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{n}-\frac{1}{n+1}$
$=1-\frac{1}{n+1}=\frac{n}{n+1}$