Problem source

\begin{questionparts}
 \item If $\mathrm{f}(r)$ is a function defined
for $r=0,1,2,3,\ldots,$ show that 
\[
\sum_{r=1}^{n}\left\{ \mathrm{f}(r)-\mathrm{f}(r-1)\right\} =\mathrm{f}(n)-\mathrm{f}(0).
\]
\item If $\mathrm{f}(r)=r^{2}(r+1)^{2},$ evaluate $\mathrm{f}(r)-\mathrm{f}(r-1)$
and hence determine ${\displaystyle \sum_{r=1}^{n}r^{3}.}$
\item Find the sum of the series $1^{3}-2^{3}+3^{3}-4^{3}+\cdots+(2n+1)^{3}.$ 
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$
\begin{align*}
&& \sum_{r=1}^n \left (f(r) - f(r-1) \right) &= \cancel{f(1)} - f(0) + \cdots\\
&&&\quad\, \cancel{\f(2)}-\cancel{f(1)} + \cdots \\
&&&\quad\, \cancel{f(3)}-\cancel{f(2)} + \cdots \\
&&&\quad\, +\cdots + \cdots \\
&&&\quad\, \cancel{f(n-1)}-\cancel{f(n-2)} + \cdots \\
&&&\quad\, f(n)-\cancel{f(n-1)} \\
&&&=f(n) - f(0)
\end{align*}

\item If $f(r) = r^2(r+1)^2$ then 
\begin{align*}
&& f(r) - f(r-1) &= r^2(r+1)^2 - (r-1)r^2 \\
&&&= r^2((r+1)^2-(r-1)^2) \\
&&&=4r^3
\end{align*}

Therefore \begin{align*}
&& \sum_{r=1}^n 4r^3 &= n^2(n+1)^2-0 \\
\Rightarrow && \sum_{r=1}^n r^3 &= \frac{n^2(n+1)^2}{4} \\
\end{align*}

\item So
\begin{align*}
&& \sum_{r=1}^{2n+1} (-1)^{r-1}r^3 &= \sum_{r=1}^{2n+1} r^3 - 2 \sum_{r=1}^n (2r)^3 \\
&&&= \frac14(2n+1)^2(2n+2)^2 - 8 \frac{n^2(n+1)^2}{4} \\
&&&= (2n+1)^2(n+1)^2 - 2n^2(n+1)^2 \\
&&&= (n+1)^2(4n^2+4n+1 - 2n^2) \\
&&&= (n+1)^2(2n^2+4n+1) 
\end{align*}
\end{questionparts}