This is just a trig identity question. That I have failed, at every turn, to figure out myself. My efforts in understanding PDE's have been thwarted by inability to do this trig manipulation. The cursed 'Simplifying, we find the solution:" has knocked another learner down.
$\mathbb The Problem:$ Using both versions of the identities (a) & (b) I still cannot see how my book got from (1) to (3) using (2). They set n=2k+1 for all odd values of n, which are what is valid for this sequence. My lack of clarity has to do with the trig identities, I cannot recreate the steps used when they plug (2) back into (1) and then 'simplify' for the results (3).
I would really like to know how they did this and what they used?
$(a)$$$ \sin^2(x)=\frac{(1-\cos 2x)}{2}$$ $(b)$$$ \sin^2(\frac{x}{2})=\frac{(1-\cos x)}{2}$$ $(1)$ $$u(x,y)=\sum_{n=0}^\infty B_n \sin\frac{n\pi}{a}x\sinh \frac{n\pi}{a}y$$ $(2)$$$B_n=\frac{200}{\sinh n\pi}\int_0^1\sin n\pi x \mathrm dx=\frac{200}{n\pi \sinh n\pi}(1-\cos n\pi)$$ $(3)$ $$u(x,y)=\frac{(400)}{\pi}\sum_{k=0}^\infty \frac{\sin(2k+1)\pi x}{(2k+1)} \frac{\sinh(2k+1)\pi y}{\sinh(2k+1)\pi}$$
Just concentrate on what $B_n$ is. If you replace, $(1-\cos(n\pi))$ by $2\sin^2(n\pi/2)$, it's easy to see that $B_n$ is zero when $n$ is even, and $$\frac{400}{n\pi\sinh(n\pi)}$$ when $n$ is odd.
Any odd number can be expressed as $2k+1$ for some appropriate integer $k$. That's all there is to it.