Question

The sum of $n$ terms of the series $\frac{\frac{1}{2}\cdot \frac{2}{2}}{1^3}+\frac{\frac{2}{2}\cdot \frac{3}{2}}{1^3+2^3}+\frac{\frac{3}{2}\cdot \frac{4}{2}}{1^3+2^3+3^3}+\dots \dots .$. is $-$

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

Answer 1

Answer: (B)

$n^{th}$ term of the given series
$T_n=\frac{\frac{n}{2}\cdot \frac{n+1}{2}}{\Sigma n^3}=\frac{\frac{1}{4}n(n+1)}{\frac{1}{4}{n}^2(n+1{)}^2}=\frac{1}{n(n+1)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)$
$\therefore S_n=\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots +\left(\frac{1}{n}-\frac{1}{n+1}\right)\right]$
$=1-\frac{1}{n+1}=\frac{n}{n+1}$