Question

If $\sum \sum _{0\le i<j\le n}j{}^n{C}_i=320$.
Then the value of $n$ is

Answer 1

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

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

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

$=\frac{n(n+1)}{2}\sum _{r=0}^{n-1}{}^{n-1}{C}_r+\frac{n(n-1)}{2}\sum _{r=1}^{n-1}{}^{n-2}{C}_{r-1}=\frac{n(n+1)}{2}{2}^{n-1}+\frac{n(n-1)}{2}{2}^{n-2}=n{2}^{n-3}(2n+2+n-1)=n(3n+1)2^{n-3}$

$\Rightarrow n(3n+1)2^{n-3}=320\Rightarrow n(3n+1)2^{n-3}=5\left(16\right)2^2\Rightarrow n=5$