$\displaystyle \Sigma [i=1, 15] i(i-1)^2$
I can see why you'd be a bit puzzled ..... here's the trick:
Substitute $\displaystyle n - 1 = m \Rightarrow n = m+1$ into $\displaystyle \sum_{n=1}^{15}n(n-1)^2$. Then you get:
$\displaystyle \sum_{m=0}^{{\color{red}14}} (m+1) m^2 = \sum_{m={\color{red}0}}^{14} (m^3+m^2)$.
Note the corrections (in red) to some typos.
Because that's what Mathstud28 did to get the result that you asked about. If you don't like that, just expand the original summand:
$\displaystyle \sum_{i=1}^{15} i (i - 1)^2 = \sum_{i=1}^{15} (i^3 - 2i^2 + i) = \sum_{i=1}^{15} i^3 - 2 \sum_{i=1}^{15} i^2 + \sum_{i=1}^{15}i$.
Now use the usual formulae on each term.