Evaluate nested summation of a function I'm trying to relearn summation simplification, I haven't touch math in a while.  I'm having trouble simplify this nested summation down and I don't even know where to start. Could anyone please give me a detail break down of what I need to do here? Thank you.

$\sum_{i=1}^n \sum_{j=i}^n c*(n-j)$
Hint:

For a fixed $i$, $1\le i\le n$ we have 
\begin{align*}
\sum_{j=i}^nc*(n-j)&amp;=c*\sum_{j=i}^n(n-j)=c\frac{(n-i)(n-i-1)}{2}=\frac{c}{2}\left[n^2-(2n-1)i+i^2\right]
\end{align*}
Then $$\sum_{i=1}^n \sum_{j=i}^nc*(n-j)=\frac{c}{2}\sum_{i=1}^n\left[n^2-(2n-1)i+i^2\right]=\frac{c}{2}\left[n^3-(2n-1)\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}\right]$$
It may be helpful to swap order of summation. Note change in index limits. 

$$\begin{align}
\sum_{i=1}^{n}\sum_{j=i}^n c(n-j)&amp;=\sum_{j=1}^n\sum_{i=1}^j c(n-j)\\
&amp;=c\sum_{j=1}^{n}j(n-j)\\
&amp;=c\left[n\sum_{i=1}^{n}j-\sum_{i=1}^n j^2\right]\\
&amp;=c\left[n\cdot \frac12n(n+1)-\frac16n(n+1)(2n+1)\right]\\
&amp;=\frac c6(n-1)n(n+1)
\end{align}$$



Alternatively:

$$\begin{align}
\sum_{i=1}^nj(n-j)&amp;=(n-1)\sum_{i=1}^nj-\sum_{i=1}^nj(j-1)\\
&amp;=(n-1)\sum_{i=1}^n\binom j1-2\sum_{i=1}^n\binom j2\\
&amp;=(n-1)\binom {n+1}2-2\binom {n+1}3\\
&amp;=3\binom {n+1}3-2\binom{n+1}3\\
&amp;=\binom{n+1}3\\
c\sum_{i=1}^nj(n-j)&amp;=c\binom{n+1}3\\
&amp;=\frac c6 (n-1)n(n+1)\end{align}$$