<s>[INST] Find the Sum using bijection Find the sum of $S=\displaystyle\sum_{i,j,k \ge 0, i+j+k=17} ijk$.
I am looking for a solution that uses some bijection.
I couldn't find any bijection.
I am able to do the problem by other method by observing that,
$S$ is the coefficient of $x^{17}$ in,
$(x+2x^2+3x^3+4x^4........)^3$ and then sum the last formal power series and then find the coefficient of  $x^{17}$ in that sum.
But the hint says find a bijection.
so, please help.
 [/INST] For this you'll require the sums up to fourth powers:
$\begin{align}
\sum_{0 \le k \le n} k
  &= \frac{n (n + 1)}{2} \\
\sum_{0 \le k \le n} k^2
  &= \frac{n (n + 1) (2 n + 1)}{6} \\
\sum_{0 \le k \le n} k^3
  &= \frac{n^2 (n + 1)^2}{4} \\
\sum_{0 \le k \le n} n^4
  &= \frac{n^2 (6 n^3 + 15 n^2 + 10 n - 1)}{30}
\end{align}$
So you want:
$\begin{align}
\sum_{\substack{i + j + k = n \\
       i, j, k \ge 0}}
         i j k
  &= \sum_{0 \le i \le n}
       i \sum_{0 \le j \le n - i}
           j  (n - i - j) \\
  &= \sum_{0 \le i \le n}
       i \sum_{0 \le j \le n - i} (n j - i j - j^2) \\
  &= \sum_{0 \le i \le n}
       i \left(
           n \frac{(n - i) (n - i + 1)}{2}
             - i \frac{(n - i) (n - i + 1)}{2}
             - \frac{(n - i) (n - i + 1) (2 n - 2 i + 1}{6}
         \right) \\
  &= \sum_{0 \le i \le n}
       \frac{i (n^3 - n) - i^2 (3 n^2 - 1) + 3 i^3 n - i^4}{6} \\
  &= \frac{(n - 1) n (3 n^3 + 3 n^2 - 12 n - 10)}{360}
\end{align}$
Phew!
Thanks to maxima for routine algebra help. Any transcription errors are mine alone.
</s>