Question

If $a_n=\sum _{r=0}^n\frac{1}{{}^n{C}_r}$, then the value of $\sum _{r=0}^n\frac{n-2r}{{}^n{C}_r}$ is

• A. $\frac{n}{2}{a}_n$
• B. $\frac{1}{4}{a}_n$
• C. $n{a}_n$
• D. $0$

Answer 1

Answer: (D)

$0$

$a_n={\sum }_{r=0}^n\frac{1}{{{}^{n}C}_r},{\sum }_{r=0}^n\frac{n-2r}{{{}^{n}C}_r}=?$

$\Rightarrow a_n=\frac{1}{{{}^{n}C}_0}+\frac{1}{{{}^{n}C}_1}+\frac{1}{{{}^{n}C}_2}+......+\frac{1}{{{}^{n}C}_{n-1}}+\frac{1}{{{}^{n}C}_n}$

$=\frac{1}{{{}^{n}C}_0}+\frac{1}{{{}^{n}C}_1}+........+\frac{1}{{{}^{n}C}_1}+\frac{1}{{{}^{n}C}_0}$

Because ${{}^{n}C}_r={{}^{n}C}_{n-r}$

$\Rightarrow a_n=1+[\frac{2}{{{}^{n}C}_1}+\frac{2}{{{}^{n}C}_2}+.....]+1$ $({{}^{n}C}_0={{}^{n}C}_n=1)$

$\Rightarrow \frac{a_n-2}{2}=[\frac{1}{{{}^{n}C}_1}+\frac{1}{{{}^{n}C}_2}+....]$

$\Rightarrow {\sum }_{n=0}^r\frac{r}{{{}^{n}C}_r}=0+\frac{1}{{{}^{n}C}_1}+\frac{2}{{{}^{n}C}_2}+.....+\frac{n-1}{{{}^{n}C}_{n-1}}+\frac{n}{{{}^{n}C}_n}$

$=\frac{1}{{{}^{n}C}_1}+\frac{2}{{{}^{n}C}_2}+.....+\frac{n-2}{{{}^{n}C}_2}+\frac{n-1}{{{}^{n}C}_1}+n$

$=[\frac{1+(n-1)}{{{}^{n}C}_1}+\frac{2+(n-2)}{{{}^{n}C}_2}+....]+n$

$=n[\frac{1}{{{}^{n}C}_1}+\frac{1}{{{}^{n}C}_2}+....]+n$

$=n\left(\frac{a_n-2}{2}\right)+n$

$\Rightarrow {\sum }_{r=0}^n\frac{r}{{{}^{n}C}_r}=n(\frac{a_n-2}{2}+1)=n\left(\frac{a_n}{2}\right)$

$\Rightarrow {\sum }_{r=0}^n\frac{n-2r}{{{}^{n}C}_r}={\sum }_{r=0}^n\frac{n}{{{}^{n}C}_r}-2{\sum }_{r=0}^n\frac{r}{{{}^{n}C}_r}$

$=n\left(a_n\right)-2\left(n\frac{a_n}{2}\right)$

$=n{a}_n-n{a}_n$

$=0.$

Hence, the answer is $0.$