New Year Maths 2016: $\sum_{r=3}^{\; 3^2}r^3=2016$

$$\large\begin{align} &\color{darkblue} {\sum_{\qquad\qquad r={\sum_{m=0}^\infty\left(\frac{n-1}n\right)^m }}^{\quad \qquad\qquad \sum_{m=0}^\infty\left(\frac{n^2-1}{n^2}\right)^m}}\color{purple}{r^n}\\\\ =&\color{darkblue}{\sum_{\qquad\qquad r={\sum_{m=0}^\infty\left(1-\frac 1n\right)^m }}^{\quad \qquad\qquad \sum_{m=0}^\infty\left(1-\frac 1{n^2}\right)^m}}\color{purple}{r^n}\\\\ =&\color{darkblue}{\sum_{\qquad \qquad r=1\big/\left[1-\left(1-\frac 1n\right)\right]}^{\quad \qquad \qquad 1\big/\left[1-\left(1-\frac 1{n^2}\right)\right]}}\color{purple}{r^n}\\\\ =&\color{darkblue}{\qquad\qquad \; \sum_{r=n}^{\; \;n^2} }\color{purple}{r^n}&&=\color{red}{(n-1)^{n+2}}\color{orange}{n^{n-1}}\color{green}{(2n+1)} \end{align}$$ By inspection, equality holds when $n=3$, giving the interesting summation result $$\large\begin{align}\color{darkblue}{\sum_{r=3}^{\; 3^2}} \color{purple}{r^3} &=\color{red}{2^5\cdot}\color{orange}{3^2\cdot}\color{green}{7}\\ \large\color{red}{\sum_{r=3}^{\; 3^2}}\color{red}{r^3} &\color{red}{=2016} \end{align}$$ Happy New Year, everyone!!