A weird Series

Before reading this solution, one might want to read up on Sum of n, n², or n³ wiki.

The sum equals to,

$S_n=\sum_{k=1}^n\left(\frac{\sum_{i=1}^k i}{k}\right)^2=\sum_{k=1}^n\left( \frac{k(k+1)/2}{k}\right)^2=\sum_{k=1}^n\left(\frac{k+1}2\right)^2$

where the last two equalities follows by sum of first $k$ positive integers formula.

Now, just shift the index. For the case of $n=16$ , we have,

$S_{16}=\sum_{k=1}^{16}\left(\frac{k+1}{2}\right)^2=\frac 14\sum_{k=2}^{17} k^2=\frac 14\left(\frac{1}{6}(17)(17+1)(34+1)-1^2\right)=\boxed{446}$

where the last two equalities follows by sum of squares formula.

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You can also write down the general result for $S_n$ which is,

$S_n=\frac 14\left(\frac{(n+1)(n+2)(2n+3)}{6}-1\right)~\forall~n\in\Bbb{Z^+}$

This general result is obtained by proceeding similarly as done for the $n=16$ case and is left as an exercise to the interested reader.

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I really like the comment in the Challenge Master note regarding the use of Nicomachus' Theorem. The series can also be represented as,

$S_n=\sum_{k=1}^n\frac{T_k^2}{k^2}$

where $T_k$ is the $k^{\textrm{th}}$ triangular number . Applying Nichomachus' Theorem, we get,

$S_n=\sum_{k=1}^n\frac{\sum_{i=1}^k i^3}{k^2}=\sum_{k=1}^n\left(\frac{k+1}{2}\right)^2$

where the last equality follows by sum of cubes formula and simple algebraic manipulation.

The rest is the same as the conclusion of the original solution.

$\begin{aligned} S_{16} & = 1 + \frac{9}{4} + \frac{36}{9}+ \frac{100}{16} + \frac{225}{25} + \frac{441}{36} + ...+ a_{16} \\ & = \left( \frac{1}{1} \right)^2 + \left( \frac{1+2}{2} \right)^2 + \left( \frac{1+2+3}{3} \right)^2 +...+\left( \frac{1+2+3+...+16}{16} \right)^2 \\ & = \sum_{k=1}^{16} {\left( \frac{\sum_{i=1}^{k} {i} }{k} \right)^2} = \sum_{k=1}^{16} {\left( \frac{k(k+1)}{2k} \right)^2} = \sum_{k=1}^{16} {\left( \frac{k+1}{2} \right)^2} \\ & = \frac {1}{4} \sum_{k=1}^{16} {\left( k^2 + 2k+ 1 \right)} = \frac {1}{4} \left( \sum_{k=1}^{16} {k^2} + 2\sum_{k=1}^{16} {k} + \sum_{k=1}^{16} {1} \right) \\ & = \frac{1}{4} \left( \frac{16(16+1)(32+1)}{6} + \frac{2(16)(16+1)}{2} + 16 \right) \\ & = \boxed{446} \end{aligned}$