Evaluating Sums $\sum_{i=1}^{n} \sum_{j=0}^{n-i}$

I'm unsure how to evaluate sums when the second sum has $n-i$ on the top.

$$\sum_{i=1}^{n} \sum_{j=0}^{n-i} (3j^{2} - 2)$$ $$=\sum_{i=1}^{n} (\sum_{j=0}^{n-i} 3j^{2} - \sum_{j=0}^{n-i}2)$$ $$=\sum_{i=1}^{n} (\sum_{j=0}^{n-i} 3(\frac {n(n+1)(n+2)} 6 ) - 2n)$$

From here I'm lost to what I could do. Could some one please explain how to evaluate this sum in detail? I'm confused.

First off, that last step isn’t right, so I’m going to start from the beginning.

\begin{align*} \sum_{i=1}^n\sum_{j=0}^{n-i}\left(3j^2-2\right)&=\sum_{i=1}^n\left(\sum_{j=0}^{n-i}3j^2-\sum_{j=0}^{n-i}2\right)\\\\ &=\sum_{i=1}^n\sum_{j=0}^{n-i}3j^2-\sum_{k=1}^n\sum_{j=0}^{n-i}2\;. \end{align*}

Now let’s deal separately with these summations. There are $n-i+1$ terms in the inner summation, so

\begin{align*} \sum_{i=1}^n\sum_{j=0}^{n-i}2&=\sum_{i=1}^n2(n-i+1)\\\\ &=2\left(\sum_{i=1}^n(n+1)-\sum_{i=1}^ni\right)\\ &=2\left(n(n+1)-\frac{n(n+1)}2\right)\\\\ &=n(n+1)\;. \end{align*}

The first one is a little messier:

\begin{align*} \sum_{i=1}^n\sum_{j=0}^{n-i}3j^2&=\sum_{i=1}^n\frac{(n-i)(n-i+1)\big(2(n-i)+1\big)}2\\\\ &=\frac12\sum_{i=1}^n\Big((n-i)(n-i+1)\big(2(n-i)+1\big)\Big)\\\\ &\overset{*}=\frac12\sum_{k=0}^{n-1}k(k+1)(2k+1)\\\\ &=\frac12\sum_{k=0}^{n-1}\left(2k^3+3k^2+k\right)\\\\ &=\sum_{k=0}^{n-1}k^3+\frac32\sum_{k=0}^{n-1}k^2+\frac12\sum_{k=0}^{n-1}k\\\\ &=\left(\frac{(n-1)n}2\right)^2+\frac{(n-1)n\big(2(n-1)+1\big)}4+\frac{(n-1)n}4\;. \end{align*}

The step marked with an asterisk is accomplished by letting $k=n-i$: as $i$ runs from $1$ through $n$, $n-i$ runs from $n-1$ down through $0$. Now just simplify this last result, combine with the first one, and simplify again to get the final result.

Realize that the top of the first sum ranges from $0$, when $i=n$, to $n-1$, when $i=1$. So by making the substitution $i=n-l$, we have $$\sum_{i=1}^{n} \sum_{j=0}^{n-i} (3j^{2} - 2)=$$ $$\sum_{l=0}^{n-1} \sum_{j=0}^{l} (3j^{2} - 2)=$$ $$\sum_{l=0}^{n-1} \left(3\frac{l(l+1)(2l+1)}{6} - 2(l+1)\right)=$$ $$\sum_{l=0}^{n-1} \left(l^3+\frac{3l^2}{2}-\frac{3l}{2}-2\right)=$$ $$\frac{n^2(n-1)^2}{4}+\frac{n(n-1)(2n-1)}{4}-\frac{3n(n-1)}{4}-2n=$$ $$\frac{n^4-5n^2-4n}{4}$$