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Decode the following summation to welcome the new year!

Find integer $n$ such that $$\large\color{darkblue}{\sum_{\qquad \qquad r={\sum_{m=0}^\infty\left(\frac{n-1}n\right)^m }}^{\qquad \qquad \quad \sum_{m=0}^\infty\left(\frac{n^2-1}{n^2}\right)^m}}\color{purple}{r^n}=\color{red}{(n-1)^{n+2}}\color{orange}{n^{n-1}}\color{green}{(2n+1)}$$


Background
Every year around new year time, questions pop up on MSE on the numerical properties of the new year (such as this, this and this), These have yielded some interesting responses. I was particularly impressed with Jack d'Aurizio's response here and tried to come up with something similar myself.

After some experimenting, I found a neat summation result for 2016. I thought it would be more interesting if formulated as a problem instead, and "enhanced" using typographically interesting elements, e.g. a summation where the limits are themselves summations. Hence the formulation of problem posted.

I posted the answer myself as the intention is to share the result. It wasn't clear if an analytical approach would work (it would be nice if someone could show that this is possible). However, a numerical approach using the first few values of $n$ will soon lead to the solution.

The "recreational mathematics" tag indicates that this is something done for fun in the spirit of good cheer for the festive period. Thanks for reading (special thanks for those who voted to reopen the question) and Happy New Year!

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  • $\begingroup$ Tried out 3, 7,8, 12 times..no avail. $\endgroup$ – Narasimham Jan 1 '16 at 17:58
  • $\begingroup$ nice...................+1@hypergeometric $\endgroup$ – Bhaskara-III Jan 2 '16 at 4:48
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$$\large\begin{align} &\color{darkblue} {\sum_{\qquad\qquad r={\sum_{m=0}^\infty\left(\frac{n-1}n\right)^m }}^{\quad \qquad\qquad \sum_{m=0}^\infty\left(\frac{n^2-1}{n^2}\right)^m}}\color{purple}{r^n}\\\\ =&\color{darkblue}{\sum_{\qquad\qquad r={\sum_{m=0}^\infty\left(1-\frac 1n\right)^m }}^{\quad \qquad\qquad \sum_{m=0}^\infty\left(1-\frac 1{n^2}\right)^m}}\color{purple}{r^n}\\\\ =&\color{darkblue}{\sum_{\qquad \qquad r=1\big/\left[1-\left(1-\frac 1n\right)\right]}^{\quad \qquad \qquad 1\big/\left[1-\left(1-\frac 1{n^2}\right)\right]}}\color{purple}{r^n}\\\\ =&\color{darkblue}{\qquad\qquad \; \sum_{r=n}^{\; \;n^2} }\color{purple}{r^n}&&=\color{red}{(n-1)^{n+2}}\color{orange}{n^{n-1}}\color{green}{(2n+1)} \end{align}$$ 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!!

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    $\begingroup$ So $$2016=3^3+4^3+5^3+6^3+7^3+8^3+9^3$$ Wonderful and Happy New Year! $\endgroup$ – mrprottolo Dec 31 '15 at 13:24
  • $\begingroup$ $$2016=\sum_{k=1}^{63} k$$ $\endgroup$ – Hexacoordinate-C Jan 1 '16 at 17:49
  • $\begingroup$ @Shadock - Yes, thanks for pointing that out! I did also notice that $$\binom {64}2=2016$$ and thought of posting a question like "find $n, y$ such that they satisfy the following: $$\binom {n^y}n=2016$$ " :) $\endgroup$ – hypergeometric Jan 1 '16 at 17:53
  • $\begingroup$ @mrprottolo - Yes! Thanks for the expansion, and Happy New Year! $\endgroup$ – hypergeometric Jan 1 '16 at 18:04
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    $\begingroup$ What it's more impressive for me it's that $2016$ is a triangular and hexagonal number ! :) $\endgroup$ – Hexacoordinate-C Jan 1 '16 at 18:06

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