Question

The value of ${\sum }_{r=1}^n{(-1)}^{r+1}\frac{{}^n{C}_r}{r+1}$ is equal to

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

Answer 1

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

we know that ${(1-x)}^n={\sum }_{r=0}^n{}^n{C}_r{(-x)}^r$
by integrating ${\int }_0^1{(1-x)}^{n}dx={\int }_0^1{\sum }_{r=0}^n{}^n{C}_r{(-1)}^r{x}^{r}dx$
$-[-\frac{1}{n+1}]={\left[{\sum }_{r=0}^n\frac{{}^n{C}_r{(-1)}^r{x}^{r+1}}{r+1}\right]}_0^1$

$\Rightarrow \frac{1}{n+1}=\frac{{}^n{C}_0{(-1)}^0}{0+1}-{\sum }_{r=1}^n\frac{{}^n{C}_r{(-1)}^{r+1}}{r+1}$
$\Rightarrow {\sum }_{r=1}^n\frac{{}^n{C}_r{(-1)}^{r+1}}{r+1}=1-\frac{1}{n+1}=\frac{n}{n+1}$
Ans: D