The Series of Madness

This is a non-conventional method.

$\displaystyle x + \frac{1}{x} = 4$

Let, $\displaystyle x = {e}^{z}$

Then,

$\displaystyle \cosh(z) = 2$

And,

$\displaystyle A = \sum_{n=1}^{50}{2\cosh(nz)}$

$\displaystyle B = \sum_{n=1}^{10}{2\cosh((5n-2)z)}$

$\displaystyle \frac{A}{B} = \frac{\displaystyle\sum_{n=1}^{50}{2\cosh(nz)}}{\displaystyle\sum_{n=1}^{10}{2\cosh((5n-2)z)}}$

Using the "Angles in AP formula" for $\cos(a)$ in terms of $\cosh(a)$ [We can do that since they are in fraction]

$\displaystyle \frac{A}{B} = \frac{\displaystyle \frac{\sinh(50z/2)\cosh(51z/2)}{\sinh(z/2)}}{\displaystyle \frac{\sinh(50z/2)\cosh(51z/2)}{\sinh(5z/2)}} = \frac{\sinh(5z/2)}{\sinh(z/2)}$

$\displaystyle \frac{A}{B} = 2\cosh(z) + 2\cosh(2z) + 1$

Substituting $\cosh(z) = 2$ ,

$\displaystyle \frac{A}{B} = \boxed{19}$

Let $a_n=x^n+\dfrac{1}{x^n}$ .

From the hypothesis, we get $x^2+1=4x$ or $x^2=4x-1$ .

Similarly, we have $\dfrac{1}{x^2}=\dfrac{4}{x}-1$ .

For integer $n\ge 3$ , we have:

$x^n=x^{n-2}. x^2=x^{n-2}(4x-1)=4x^{n-1}-x^{n-2}$

$\dfrac{1}{x^n}=\dfrac{1}{x^{n-2}}.\dfrac{1}{x^2}=\dfrac{1}{x^{n-2}}\left(\dfrac{4}{x}-1\right)=\dfrac{4}{x^{n-1}}-\dfrac{1}{x^{n-2}}$

Thus, $a_n=4a_{n-1}-a_{n-2}$ for all integer $n\ge3$ .

For each integer $n\ge 3$ , we have:

$\quad a_{n+2}+a_{n+1}+a_n+a_{n-1}+a_{n-2}$

$=4a_{n+1}-a_n+a_{n+1}+a_n+a_{n-1}+4a_{n-1}-a_n$

$=5(a_{n+1}+a_{n-1})-a_n$

$=20a_n-a_n=19a_n$ .

Thus,

$\displaystyle A=\sum_{n=1}^{10} \left(a_{5n}+a_{5n-1}+a_{5n-2}+a_{5n-3}+a_{5n-4}\right)=\sum_{n=1}^{10} 19a_{5n-2}=19B$

Or $\dfrac{A}{B}=\boxed{19}$ .

My favorite solution would be this one. Notice that, for all $a, m, n \in Z$ :

$(a^m+\frac{1}{a^m})(a^n+\frac{1}{a^n})=a^{m+n}+a^{m-n}+\frac{1}{a^{m-n}}+\frac{1}{a^{m+n}}$

With this, then:

$(a^m+\frac{1}{a^m})\displaystyle \sum_{i=-n}^n a^i=\displaystyle \sum_{m-n}^{m+n}(a^i+\frac{1}{a^i})$

By expanding $B$ , multiplying it by $\displaystyle \sum_{i=-2}^2 x^i$ , and shrinking the value back to sum notation form, we get:

$(x^2+x+1+\frac{1}{x}+\frac{1}{x^2})B = \displaystyle \sum_{n=1}^{10}(\displaystyle \sum_{k=-4}^0 (x^{5n+k}+\frac{1}{x^{5n+k}}))=\displaystyle \sum_{n=1}^{50}(x^n+\frac{1}{x^n})=A$

Thus, $\frac{A}{B}=x^2+x+1+\frac{1}{x}+\frac{1}{x^2}=(x+\frac{1}{x})^2+(x+\frac{1}{x})-1=16+4-1=19$ .