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)&=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.
$$\begin{align} \sum_{i=1}^{n}\sum_{j=i}^n c(n-j)&=\sum_{j=1}^n\sum_{i=1}^j c(n-j)\\ &=c\sum_{j=1}^{n}j(n-j)\\ &=c\left[n\sum_{i=1}^{n}j-\sum_{i=1}^n j^2\right]\\ &=c\left[n\cdot \frac12n(n+1)-\frac16n(n+1)(2n+1)\right]\\ &=\frac c6(n-1)n(n+1) \end{align}$$
Alternatively:
$$\begin{align} \sum_{i=1}^nj(n-j)&=(n-1)\sum_{i=1}^nj-\sum_{i=1}^nj(j-1)\\ &=(n-1)\sum_{i=1}^n\binom j1-2\sum_{i=1}^n\binom j2\\ &=(n-1)\binom {n+1}2-2\binom {n+1}3\\ &=3\binom {n+1}3-2\binom{n+1}3\\ &=\binom{n+1}3\\ c\sum_{i=1}^nj(n-j)&=c\binom{n+1}3\\ &=\frac c6 (n-1)n(n+1)\end{align}$$
