Why do we use 2n-1 instead of n in this Fourier series?

[fufufuha]

http://www4.okfoto.co.kr/S_storage4/314500/A05120618494148_t.jpg http://www4.okfoto.co.kr/S_storage4/314500/A05120618494175_t.jpg
It's solution is http://www4.okfoto.co.kr/_Inc/Process/proSendbinaryHow.asp?w=400&h=400&path=\\100.100.100.11\link_album\pic_root\storage4\314500\A05120618494172.jpg
What I know is that insert 2n-1 instead of n
please... explain to me
I don't know this reason

 

[Integral]

Please make an effort to provide a more complete statement of your problem.

 

[fufufuha]

First I'm sorry that I can't write english well.
I have a problem's solution.
But I don't understand solution
For example Another problem's solution is n instead of 2n-1
But this problem's solution uses 2n-1 instead of n
I understood that 2n-1 is odd
but why do 2n-1 instead of n ?

 

[HallsofIvy]

When x= 1 or -1, 1-|x|= 0. Suppose you had just $\frac{n\pi}{2}x$. Then for even values of n, x= 1, you would have $cos(m\pi)$ (m= n/2) which is either -1 or 1. By restricting the numerator to be odd, you are making sure that you always have, for x= 1, $cos(\frac{\pi}{2})$, $cos(\frac{3\pi}{2})$, etc.

By the way, how does one do Fourier Series in precalculus?