Prove that $\Sigma_{i=0}^{n-1} (\Sigma_{j=i+1}^n(j(^nC_i) + i(^nC_j))) = n^22^{n-1}$ I have tried setting
$$P=\Sigma_{i=0}^{n-1} (\Sigma_{j=i+1}^n(j(^nC_i) + i(^nC_j))$$
then
$$P=\Sigma_{i=0}^{n-1} (\Sigma_{j=i+1}^n((n-j)(^nC_i) + (n-i)(^nC_j))$$
now adding them both
$$2P=\Sigma_{i=0}^{n-1} (\Sigma_{j=i+1}^n(n((^nC_i)+(^nC_j))))$$
I think that we have to go ahead from here but I'm not able to think of anything else. Am I even on the right path? Please Help. Thanks!
 A: 
We obtain
\begin{align*}
\color{blue}{\sum_{i=0}^{n-1}}&\color{blue}{\sum_{j=i+1}^n\left(j\binom{n}{i}+i\binom{n}{j}\right)}\\
&=\sum_{0\leq i<j\leq n}\left(j\binom{n}{i}+i\binom{n}{j}\right)\tag{1}\\
&=\sum_{0\leq i\ne j\leq n}i\binom{n}{j}\tag{2}\\
&=\sum_{0\leq i,j\leq n}i\binom{n}{j}-\sum_{0\leq i=j\leq n}i\binom{n}{j}\\
&=\left(\sum_{i=0}^ni\right)\left(\sum_{j=0}^n\binom{n}{j}\right)-\sum_{i=0}^ni\binom{n}{i}\tag{3}\\
&=\frac{1}{2}n(n+1)2^n-n\sum_{i=1}^n\binom{n-1}{i-1}\tag{4}\\
&=\frac{1}{2}n(n+1)2^n-n\sum_{i=0}^{n-1}\binom{n-1}{i}\tag{5}\\
&=\frac{1}{2}n(n+1)2^n-n2^{n-1}\\
&\,\,\color{blue}{=n^22^{n-1}}
\end{align*}
and the claim follows.

Comment:

*

*In (1) we write the index region somewhat more conveniently to better see what's going on.


*in (2) we use the symmetry of the index region and the summands.


*In (3) we can write the sums separating the indices $i$ and $j$.


*In (4) we simplify using closed forms and apply the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.


*In (5) we shift the index to start with $i=0$ and use again a closed form expression as in (4).
A: I'm going ahead and assume $^nC_i=\binom{n}{i}$ and so
$$2P=n\sum_{i=0}^n \sum_{j=i+1}^n \left( \binom{n}{i} + \binom{n}{j} \right) = n\sum_{i=0}^n (n-i)\binom{n}{i} + n\sum_{j=0}^n\sum_{i=0}^{j-1} \binom{n}{j}
=n^2\sum_{i=0}^{n} \binom{n}{i}=n^2 2^n$$
