Solution to a Madoka Magica Problem

In chapter one of the anime series Maho Shojo Madoka Magica (roughly minute 13), the following problem is posed for middle school students as a whiteboard excercise:

Given $f(n) = \displaystyle{\frac{4n + \sqrt{4n^2 - 1}}{\sqrt{2n+1} + \sqrt{2n-1}}}$ find the sum of $f(1) + f(2) + f(3) + \dots + f(60)$

Akemi Homura, one of the main characters in the series, seems to solve it with some little algebraic mistakes here and there and following a fairly unconventional (although quite clever!) approach. I ended up thinking about this problem (as well as some other math problems shown in ths series and others) so much as to use it as a common excercise for my own students. So I wanted to share here the solution that (based on the calculations shown briefly on the series) I think is closest to the one Homura does:

Define $a_n = \sqrt{2n-1}$, that way $a_{n+1} = \sqrt{2(n+1)-1} = \sqrt{2n+1}$.

Note that $a_n \cdot a_{n+1} = \sqrt{(2n-1)(2n+1)} = \sqrt{4n^2 - 1}$

Now, we can write $4n$ as $(2n-1) + (2n + 1) = a_n^2 + a_{n+1}^2$ and the function $f(n)$ can be written in terms of this sequence as follows:

$f(n) = \displaystyle{\frac{a_n^2 + a_{n+1}^2 + a_n \cdot a_{n+1}}{a_n + a_{n+1}} = \frac{a_{n+1}^2 + a_n \cdot a_{n+1} + a_n^2}{a_{n+1} + a_n}}$

Multiplying up and down by $a_{n+1} - a_n$ and using the identities for difference of squares and cubes yields:

$f(n) = \displaystyle{\frac{a_{n+1}^3 - a_n^3}{a_{n+1}^2 - a_n^2}}$

As a last simplification, we note that $a_{n+1}^2 - a_n^2 = (2n+1) - (2n-1) = 2$ so the sum turns into:

$\displaystyle{\sum_{k=1}^{60} \frac{a_{k+1}^3 - a_k^3}{2}} = \frac{1}{2} \sum_{k=1}^{60} a_{k+1}^3 - a_k^3$

Which is a telescoping sum that evaluates to $a_{61}^3 - a_1^3 = (2 \cdot 61 - 1)^{3/2} - (2 \cdot 1 - 1)^{3/2} = 11^3 - 1 = 1330$

So multiplying by $\frac{1}{2}$ we have the final result:

$f(1) + f(2) + f(3) + \dots + f(60) = 665$