I'm looking to solve the following when $T_0$ is a constant:
$$\sum_{n=1}^\infty a_n \sin\left(\frac{n \pi}{2}\right)=T_0$$
If it matters this was reached from the following:
$$T(x,t)=\sum_{n=1}^{\infty}a_n \sin\left(\frac{n \pi x}{2d}\right)e^{\frac{-n^2\pi^2k^2t}{4d^2}}$$ $$T(d,0)=T_0$$
The original problem that was posted has no unique solution.
For example, letting $a_1=T_0$ and $a_n=0$ for $n\ne 1$ provides one solution. Letting $a_3=-T_0$ and $a_n=0$ for $n\ne 3$ provides another. There are an infinite number of possible solutions.
Thus, either there isn't a unique one or the problem is mis-stated.
If it has been mis-stated, then let's start by assuming that
$$T(x,0) =\sum_{n=1}^{\infty} a_n \sin(n\pi x/2d)$$
Now, let's see how we can arrive at an answer.
To that end, multiply both sides of the equation by $\sin(m\pi x/2d)$ and integrate from $0$ to $2d$.
$$\int_0^{2d} \sin(m\pi x/2d) \sum_{n=1}^{\infty} a_n \sin(n\pi x/2d) dx=T_0\int_0^{2} \sin(m\pi x/2d) dx$$
Formally, interchange the order of integration and summation and exploit the orthogonality of ${\sin (n\pi x/2d)}$ on $0\le x\le 2d$ to reveal that
$$a_m=(T_0/d)\int_0^{2d} \sin(m\pi x/2d) dx=T_0 \frac{1-(-1)^m}{(m\pi/2)}$$