Triple finite sum $\displaystyle \sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^c f(i,j,k)$ where a,b,c are fixed natural numbers and assuming $f(i,j,k)=i+j+k$. How do we calculate that sum? I mean is there any type for that sum? Function $f$ includes $i, j$ and $k$ and that  confuses me.
 A: $$\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^cf(i,j,k)=\sum_{n=3}^{a+b+c}n\sum_{i+j+k=n}1=\sum_{n=3}^{a+b+c}n\binom{n-1}{2}$$
$3\leq f(i,j,k)=i+j+k=n\leq a+b+c$ 
A: Since
$$
\sum_{i=1}^ai=\binom{a+1}{2}
$$
we have
$$
\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^ci=\binom{a+1}{2}bc
$$
Therefore,
$$
\begin{align}
\sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^c (i + j + k)
&=\left(\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^ci\right)
+\left(\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^cj\right)
+\left(\sum_{i=1}^a\sum_{j=1}^b\sum_{k=1}^ck\right)\\[6pt]
&=\quad\overbrace{\binom{a+1}{2}bc}^{\text{contribution of }i}
\quad+\quad\overbrace{a\binom{b+1}{2}c}^{\text{contribution of }j}
\quad+\quad\overbrace{ab\binom{c+1}{2}}^{\text{contribution of }k}\\[18pt]
&=\frac{abc(a+b+c+3)}{2}
\end{align}
$$
A: You want to evaluate $\sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^c (i+j+k)$  Start from the inner sum.  The $i+j$ part is constant, as only $k$ is varying, so just contributes $c(i+j)$  because there are $c$ terms.  The $k$ part gives the triangle number $\frac 12c(c+1)$, so $$\sum_{i=1}^a \sum_{j=1}^b \sum_{k=1}^c (i+j+k)=\sum_{i=1}^a \sum_{j=1}^b c(i+j)+\frac 12c(c+1)$$  Now do the sum over $j$ the same way-any term that doesn't include a $j$ is a constant and just gets multiplied by $b$ as there are $b$ of them.  The term including $j$ gives a triangle number-which one?
