Question

Find the sum $\sum _{0\le i<}\sum _{j\le n}{j}^n{C}_i$.

• A. $n(n+2)2^{n-3}$
• B. $n(n+3)2^{-3}$
• C. $n(n+2)2^{n-1}$
• D. $n(n+3)2^{n-3}$

Answer 1

Answer: (D)

$n(n+3)2^{n-3}$

$\sum _{o\le i<}\sum _{j\le n}{j}^n{C}_i$
$=\sum _{r=0}^{n-1}{}^n{C}_r\left[\right(r+1)+(r+2)+...+(n\left)\right]$
$=\sum _{r=0}^{n-1}{}^n{C}_r\left[\right(r+1)+(r+2)+...+(n\left)\right]$
$=\sum _{r=0}^n{}^n{C}_r\left(\frac{n+1}{2}\right(n-r)-\frac{r(n-r)}{2})$
$=\frac{n+1}{2}\sum _{r=0}^n(n-r){}^n{C}_r-\frac{n}{2}$
${\sum }_{r=0}^n{r}^n{C}_r+\frac{1}{2}{\sum }_{r=0}^n{r}^2$
$=\frac{n+1}{2}\sum _{r=0}^{n}r{}^n{C}_r-\frac{n}{2}\sum _{r=0}^{n}r{}^n{C}_r+\frac{1}{2}\sum _{r=0}^n{r}^2{}^n{C}_r$
$=\frac{1}{2}(\sum _{r=0}^{n}r{}^n{C}_r+\sum _{r=0}^n{r}^2{}^n{C}_r)$
$=\frac{1}{2}(n{2}^{n-1}+n(n-1)2^{n-2}+n{2}^{n-1})$
$=n(n+3)2^{n-3}$