How do you show $n^3$ as a sum of terms?

Question

We know that $n^2 = \displaystyle\sum_{i=1}^n 2i-1$.

Is there a similar way to represent $n^3$ as: $\displaystyle\sum_{i=1}^n ?$, where we replace the question mark with a term?

Answer 1

Thes comments you got so far should make for a good way through for you. Alternatively (similarly), you may look at this link to see that

\begin{align} S_1&=\sum_1^n i =\frac12n^2+\frac12n\\ S_2&=\sum_1^n i^2 =\frac13n^3+\frac12n^2+\frac16n\\ \end{align} Now look at $S_2-S_1$: \begin{align} S_2-S_1&=\sum_1^n i^2-\sum_1^n i\\ &=\frac13n^3+\frac12n^2+\frac16n-\frac12n^2-\frac12n\\ &=\frac13n^3-\frac13n\\ \end{align} Therefore \begin{align} 3\sum_1^n i^2-3\sum_1^n i&=n^3-n\\ \end{align} Now recall that $\sum_1^n 1=n$ hence \begin{align} 3\sum_1^n i^2-3\sum_1^n i&=n^3-\sum_1^n 1\\ \end{align} or \begin{align} 3\sum_1^n i^2-3\sum_1^n i+\sum_1^n 1&=n^3\\ \sum_1^n(3 i^2-3 i+ 1)&=n^3\\ \end{align}