Question

The sum of $\frac{\frac{1}{2}\times \frac{2}{2}}{1^3}+\frac{\frac{2}{2}\times \frac{3}{2}}{1^3+2^3}+\frac{\frac{3}{2}\times \frac{4}{2}}{1^3+2^3+3^3}+.....$ upto n terms is equal to

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

Answer 1

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

Solve by putting different values of n.
For n = 1, Given sum = $\frac{1}{2}$
Only option (b) is equal to $\frac{1}{2}$ for n = 1, hence is correct answer.

Conventional Approach:
The general term is
$T_n=\frac{\frac{n}{2}.\frac{n+1}{2}}{1^3+2^3+3^3+...+n^3}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\therefore S_n=1-\frac{1}{n+1}=\frac{n}{n+1}$