Question

$\sum _{r=1}^n\frac{r}{\mathrm{1.3.5.7.....}\left(2r+1\right)}$ is equal to

• A. $\frac{1}{2}\left[1-\frac{1}{\mathrm{1.3.5....}\left(2n+1\right)}\right]$
• B. $\frac{1}{4}\left[1-\frac{1}{\mathrm{1.3.5......}\left(2n-1\right)}\right]$
• C. $\frac{1}{4}\left[1+\frac{1}{\mathrm{1.3.5.....}\left(2n-1\right)}\right]$
• D. none of these

Answer 1

The correct option is A $\frac{1}{2}\left[1-\frac{1}{\mathrm{1.3.5....}\left(2n+1\right)}\right]$

$={\sum }_{r=1}^n\frac{r}{\mathrm{1.3.5.........}(2r+1)}$

$=\frac{1}{2}{\sum }_{r=1}^n\frac{2r}{\mathrm{1.3.5.........}(2r+1)}$

$=\frac{1}{2}{\sum }_{r=1}^n[\frac{1}{\mathrm{1.3.5.........}(2r-1)}-\frac{1}{\mathrm{1.3.5.........}(2r-1)(2r+1)}]$

$=\frac{1}{2}[\frac{1}{1}-\frac{1}{1.3}+\frac{1}{1.3}-\frac{1}{\mathrm{1.3.5}}+\frac{1}{\mathrm{1.3.5}}...........-\frac{1}{\mathrm{1.3.5.........}(2n+1)}]$

$=\frac{1}{2}[1-\frac{1}{\mathrm{1.3.5.........}(2n+1)}]$