Question

Find the sum of ${\sum }_{0\le i\le j\le n}\sum j\stackrel{n}{\underset{i}{C}}$

Answer 1

${\sum }_{0\le i\le j\le n}\sum j{}^n{C}_i$

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

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

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

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

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

$\Rightarrow \frac{1}{2}(n{2}^{n-1}+n(n-1)2^{n-2}+n{2}^{n-1})$

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

Therefore, $\Rightarrow {\sum }_{0\le i\le j\le n}\sum j{}^n{C}_i=n(n+3)2^{n-3}$