Do You Love Summations?

Recall that the sum of the first $n$ perfect squares is $1^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac16 n(n+1)(2n+1)$ . (reference: sum of n, n², or n³ ).

$\begin{aligned} & = \displaystyle \sum_{n=1}^{101} \left( \frac{2n+1}{\displaystyle \sum_{i=1}^{n} i^{2}} \right ) \\ & = \displaystyle \sum_{n=1}^{101} \left( \frac{2n+1}{\frac{n(n+1)(2n+1)}{6}} \right ) \\ & = 6 \displaystyle \sum_{n=1}^{101} \frac{1}{n(n+1)} \\ & = 6 \displaystyle \sum_{n=1}^{101}\left( \frac{1}{n}-\frac{1}{n+1}\right) \\ & = 6\left( \displaystyle \sum_{n=1}^{101}\frac{1}{n}- \displaystyle \sum_{n=2}^{102} \frac{1}{n} \right) \\ & = 6\left(1-\frac{1}{102}\right) \\ & = \frac{101}{17} \\ & \Rightarrow 101+17 =\boxed{118} \end{aligned}$