Question

The sum of $\frac{\frac{1}{2}.\frac{2}{2}}{1^3}+\frac{\frac{2}{2}.\frac{3}{2}}{1^3+2^3}+\frac{\frac{3}{2}.\frac{4}{2}}{1^3+2^3+3^3}+....$ 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 $\frac{n}{n+1}$

$\begin{array}{c}\frac{\frac{1}{2}.\frac{2}{2}}{1^3}+\frac{\frac{2}{2}.\frac{3}{2}}{1^3+2^3}+\frac{\frac{3}{2}.\frac{4}{2}}{1^3+2^3+3^3}+........+\frac{\frac{n}{2}\left(\frac{n+1}{2}\right)}{1^3+2^3+.......+n^3}\\ So,\\ T_r=\frac{\frac{r}{2}\left(\frac{r+1}{2}\right)}{1^3+2^3+.......+r^3}\Rightarrow \frac{\frac{1}{4}r\left(r+1\right)}{{\left[r\frac{\left(r+1\right)}{2}\right]}^2}=\frac{\frac{r\left(r+1\right)}{4}}{\frac{r{\left(r+1\right)}^2}{4}}\\ T_r=\frac{r+1-r}{r\left(r+1\right)}=\frac{1}{r}-\frac{1}{r+1}\\ S_n={\sum }_{r=1}^n{T}_r\\ ={\sum }_{r=1}\frac{1}{r}-\frac{1}{r+1}\\ =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{n}-\frac{1}{n+1}\\ S_n=\frac{1}{1}-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}\end{array}$