Fourier series help please

Hey dudes I need help with this fourier series

The odd function of f has a period of 2, it is designed in the interval 0 find the fourier series of f(x) =4x(1-x)

Thanks for your help dudes :-)

We need a Fourier sine series, since we are dealing with the odd periodic extension of f.

So, all we need to compute are b(n):
b(n) = (2/1) ∫(x = 0 to 1) 4x(1 - x) sin(nπx/1) dx
.......= 8 ∫(x = 0 to 1) (x - x^2) sin(nπx) dx
.......= 8 [-(x - x^2) cos(nπx)/(nπ) + (1 - 2x) sin(nπx)/(nπ)^2 - 2 cos(nπx)/(nπ)^3] {for x = 0 to 1}
.......= 8 [-2 cos(nπ)/(nπ)^3 + 2/(nπ)^3]
.......= 16 [1 - (-1)^n] /(nπ)^3
.......= 32/(nπ)^3, if n is odd, and 0 otherwise.

Writing n = 2k + 1 for k = 0, 1, 2, ...,
b(2k+1) = 32/((2k+1)π)^3.

Hence, the desired Fourier (sine) series is
Σ(k = 0 to ∞) [32/((2k+1)π)^3] sin((2k+1)πx).

I hope this helps!