Converting Σ i(i + 1) into a formula, given this hint The given summation is: $$\sum_{i=1}^n i(i+1)$$
The goal is to convert it into a formula which only uses n.
Solving this, I got the answer:
$$\frac{n}{3}(n+1)(n+2)$$
However, I don't believe the way I solved it is the intended method, despite arriving at the correct solution. This is because a hint is provided:
$$i(i+1) = \frac{(i+1)^3 - i^3 - 1}{3}$$
My solution did not benefit from this hint, thus I believe my method of solving was unintended.
Based on the hint, what method should be used to convert the summation into a formula?
 A: For fixed $i$, 
$\begin{eqnarray}
\frac{(i+1)^3-i^3-1}{3}&=&\frac{i^3+3i^2+3i+1-i^3-1}{3}\\
&=&\frac{3i^2+3i}{3}\\
&=&i(i+1)
\end{eqnarray}$
So
$\begin{eqnarray}
\sum_{i=1}^ni(i+1)&=&\sum_{i=1}^n\frac{(i+1)^3-i^3-1}{3}\\
&=&\frac{1}{3}\sum_{i=1}^n[(i+1)^3-i^3-1]\\
&=&\frac{1}{3}\left(\sum_{i=1}^n(i+1)^3-\sum_{i=1}^ni^3-\sum_{i=1}^n1\right)\\
&=&\frac{1}{3}\left(\sum_{i=2}^{n+1}i^3-\sum_{i=2}^ni^3-1-\sum_{i=1}^n1\right)\\
&=&\frac{1}{3}((n+1)^3-1-n)\\
&=&\frac{1}{3}(n^3+3n^2+2n)\\
&=&\frac{n(n+1)(n+2)}{3}
\end{eqnarray}$
Another approach is
$\sum_{i=1}^ni(i+1)=\sum_{i=1}^ni^2+i=\sum_{i=1}^ni^2+\sum_{i=1}^ni=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$ and simplifies.
A: $\sum_{k=1}^n k(k+ 1) = \sum_{k=1}^n \dfrac{(i+1)^3-i^3-1}3=$
$1/3[\sum_{k=1}^n (i+1)^3- \sum_{k=1}^n i^3 - \sum_{k=1}^n1]=$
$1/3[\sum_{k=2}^{n+1} i^3- \sum_{k=1}^n i^3 - n]=$
$1/3[(n+1)^3- 1^3 - n]=$
$1/3[n^3 + 3n^2 + 3n - n]=$
$1/3[n^3 + 3n^2 + 2n]=$
$n/3[n^2 + 3n + 2]=$
$\dfrac{n(n+1)(n+2)} 3$
A: The hint is incorrect, as you can easily tell by epanding the RHS.  They probably meant
$$i(i+1)=\frac{(i+1)^3-i^3-1}{3}\ ,$$
and you can use this to evaluate the result as a telescoping sum:
$$\eqalign{\sum_{i=1}^n i(i+1)
  &=\frac13\sum_{i=1}^n((i+1)^3-i^3-1)\cr
  &=\frac13((2^3+\cdots+n^3+(n+1)^3)-(1^3+2^3+\cdots+n^3)-(1+2+\cdots+n))\cr
  &=\frac13((n+1)^3-1-\frac12n(n+1))\cr}$$
and so on.
