Problem source

It is given that $\sum\limits_{r=-1}^ {n} r^2$  can be written  in the form $pn^3 +qn^2+rn+s\,$, where $p\,$, $q\,$, $r\,$ and $s$ are numbers. By setting $n=-1$, $0$, $1$  and $2$, obtain four  equations that must be satisfied by $p\,$, $q\,$, $r\,$ and $s$ 
and hence show that
\[
{ \sum\limits_{r=0} ^n} r^2= {\textstyle \frac16} n(n+1)(2n+1)\;.
\]
Given that $\sum\limits_{r=-2}^ nr^3$  can be written  in the form  $an^4 +bn^3+cn^2+dn +e\,$, show similarly that
\[
{ \sum\limits_{r=0} ^n} r^3= {\textstyle \frac14} n^2(n+1)^2\;.
\]

Solution source

\begin{align*}
n = -1: && (-1)^2 &= s - r+q -p \\
n = 0: && 1 + 0 &= s \\
n = 1: && 1 + 1 &= s + r + q + p \\
n = 2: && 2 + 2^2 &= s + 2r + 4q + 8p \\
\Rightarrow &&& \begin{cases}
1 &= s \\
1 &= s - r + q - p \\
2 &= s + r + q + p \\
6 &= s + 2r + 4q + 8p
\end{cases} \\
\Rightarrow && s &= 1 \\
&& q &= \frac12 \\
&&& \begin{cases} \frac12 &= r + p \\
3 &= 2r + 8p \end{cases} \\
\Rightarrow && r &= \frac16 \\
&& p &= \frac13 \\
\Rightarrow && \sum_{r=0}^n r^2 &= 1 + \frac16 n + \frac12 n^2 + \frac13 n^3 - (-1)^2 \\
&&&= \frac{n}{6} \l 1 + 3n + 2n^2 \r \\
&&&= \frac{n(n+1)(2n+1)}{6}
\end{align*}

Similarly,

\begin{align*}
n = -2: && (-2)^3 &= e - 2d + 4c - 8b + 16a \\
n = -1: && -8 + (-1)^3 &= e -d+c-b+a \\
n = 0: && -9 + 0^3 &= e \\
n = 1: && -9 + 1^3 &= e+d+c+b+a \\
n = 2: && -8 + 2^3 &= e+2d+4c+8b+16a \\
\Rightarrow &&& \begin{cases}
-9 &= e \\
-9 &= e - d+c -b + a \\
-8 &= e +d+c+b+a \\
-8 &= e-2d+4c-8b+16a \\
0 &= e+2d+4c+8b+16a \\
\end{cases} \\
\Rightarrow && e &= -9 \\
\Rightarrow &&& \begin{cases}
1 &=  2c+2a \\
10 &= 8c+32a \\
1 &= 2d+2b \\
8 &= 4d+16b \\
\end{cases} \\
\Rightarrow && a &= \frac14 \\
&& c &= \frac14 \\
&& b &= \frac12 \\
&& d &= 0 \\
\\
\Rightarrow && \sum_{r=0}^n r^3 &= -9 + \frac14n^2 + \frac12 n^3+\frac14 n^4 -((-1)^3+(-2)^3) \\
&&&= \frac14n^2 \l1 + 2n+n^2\r \\
&&&= \frac{n^2(n+1)^2}{4}
\end{align*}

as required