Fourier sine series for $f(x) = x(1-x)$ on the interval $[0,1]$

Question

So I'm trying to find the Fourier series for $f(x) = x(1 - x)$ on the interval $[0, 1]$ and so the coefficients are $$ b_n = \int_{0}^{1} x(1-x)\sin(n\pi x)dx $$ which is a pretty simple integration parts, from which I get $$ b_n = \frac{4\left(1-\left(-1\right)^{n}\right)}{n^{3}\pi^{3}} $$ Which would give me the association $$ x(1-x) = \frac{4}{\pi^3}\sum_{n=1}^{\infty}\frac{\left(1-\left(-1\right)^{n}\right)}{n^{3}}\sin\left(n\pi x\right) $$ But I know that the answer I'm supposed to get is $$ x(1-x) = \frac{8}{\pi^3}\sum_{n=1}^{\infty}\frac{2}{\left(2n-1\right)^{3}}\sin\left(\left(2n-1\right)\pi x\right) $$ Am I going wrong somewhere? I followed all the steps correctly and I am pretty sure I have the correct coefficients. Just a bit confused how to get the right answer

Answer 1

You are correct. Simply note that $1-(-1)^n$ is zero if $n$ is even and $2$ if $n$ is odd. An odd number can be written as $2m-1$ and finally $n$ can be substituted to $m$.