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)=4n+4n212n+1+2n1f(n) = \displaystyle{\frac{4n + \sqrt{4n^2 - 1}}{\sqrt{2n+1} + \sqrt{2n-1}}} find the sum of f(1)+f(2)+f(3)++f(60)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 an=2n1a_n = \sqrt{2n-1}, that way an+1=2(n+1)1=2n+1a_{n+1} = \sqrt{2(n+1)-1} = \sqrt{2n+1}.

Note that anan+1=(2n1)(2n+1)=4n21a_n \cdot a_{n+1} = \sqrt{(2n-1)(2n+1)} = \sqrt{4n^2 - 1}

Now, we can write 4n4n as (2n1)+(2n+1)=an2+an+12(2n-1) + (2n + 1) = a_n^2 + a_{n+1}^2 and the function f(n)f(n) can be written in terms of this sequence as follows:

f(n)=an2+an+12+anan+1an+an+1=an+12+anan+1+an2an+1+anf(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 an+1ana_{n+1} - a_n and using the identities for difference of squares and cubes yields:

f(n)=an+13an3an+12an2f(n) = \displaystyle{\frac{a_{n+1}^3 - a_n^3}{a_{n+1}^2 - a_n^2}}

As a last simplification, we note that an+12an2=(2n+1)(2n1)=2a_{n+1}^2 - a_n^2 = (2n+1) - (2n-1) = 2 so the sum turns into:

k=160ak+13ak32=12k=160ak+13ak3\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 a613a13=(2611)3/2(211)3/2=1131=1330a_{61}^3 - a_1^3 = (2 \cdot 61 - 1)^{3/2} - (2 \cdot 1 - 1)^{3/2} = 11^3 - 1 = 1330

So multiplying by 12\frac{1}{2} we have the final result:

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

Note by Benja Vera
1 year, 1 month ago

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Is there any way i can share this on facebook? i find it hilarious and, at the same time, fairly interesting

Andrea Uribe - 1 year, 1 month ago

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ohhh I can't find a share button anywhere:( But you can paste the url if you want!

Benja Vera - 1 year, 1 month ago

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Nice!, @Benja Vera

A Former Brilliant Member - 1 year, 1 month ago

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