Hello, soyeahiknow!
It is a major theorem that:
. . $\displaystyle \displaystyle{S \;=\; \sum^{\infty}_{n=1}\frac{1}{n^2} \;=\;\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots \;=\;\frac{\pi^2}{6} }$
Use this to calculate: .$\displaystyle [1]\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots$
We have: .$\displaystyle X \;=\;\dfrac{1}{2^2} + \dfrac{1}{4^2} + \dfrac{1}{6^2} + \hdots $
. . . . . . . .$\displaystyle X \;=\;\dfrac{1}{2^2}\underbrace{\bigg[\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \hdots \bigg]}_{\text{This is }S} $
. . . . . . . .$\displaystyle X \;=\;\dfrac{1}{4}\cdot\dfrac{\pi^2}{6}$
. . . . . . . .$\displaystyle X \;=\;\dfrac{\pi^2}{24}$
$\displaystyle [2]\;\dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \hdots$
Are the denominators primes? .$\displaystyle 2, 3, 5, 7, 11, 13,\:\hdots$
or a Fibonacci-type sequence? .$\displaystyle 2, 3, 5, 8, 13, 21,\:\hdots$
Prime Zeta Function -- from Wolfram MathWorld
id:A085548 - OEIS Search Results
EDIT: I should have added some comment to be clear; it's likely that tonio is right that the problem is misstated; nevertheless, if it is correctly stated then the above links are relevant.
In post #2, Soroban factored out (1/2)^2 because it evenly divides each term, and because doing so allows us to express the sum as a product of already known quantities.
For this problem, think about what happens when you add this series to the series that Soroban solved.
if I add this series to the one Soroban solves, it would just be back to the original series right?
OOOOO so all I have to do is ( original series - series with all the even numbers)= series for problem 2 ( the odd numbers). So (pi^2)/6-(pi^2)/24= (pi^2)/8
and what kind of problems are these called? Thanks so much and sorry for the typo in the beginning! I posted it at like 3 am.