I have the following problem: $$ \begin{cases} u_{xx}+u_{yy}=0&x^2+y^2<4,\quad y>0\\ u(x,y)=g(x,y)&x^2+y^2=4,\quad y>0\\ u(x,0)=u(-x,0)&\leq x\leq 2\\ u_{y}(x,0)=-u_{y}(-x,0)&\leq x\leq 2 \end{cases} $$ now, I know that for Laplace for the inside circle my answer should be in the form of the following: $$u(r,\theta)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n\left(a_n\cos(n\theta)+b_n\sin(n\theta)\right)$$
and I got: $$u(r,\theta)=\frac{A_0}{2}+\sum_{n=1}^{\infty}\left(\frac{r}{2}\right)^n\left(A_n\cos(n\theta)+B_n\sin(n\theta)\right)$$
but the official answer: $$u(r,\theta)=\frac{A_0}{2}+\sum_{n=1}^{\infty}\left(\frac{r}{2}\right)^{2n}\left(A_n\cos(2n\theta)+B_n\sin(2n\theta)\right)$$ why the $n$ became $2n$? it also appears in the Fourier coefficients at the official answer: $$ A_n=\frac{2}{\pi}\int_{0}^{\pi}h(\theta)\cos{2n\theta}d\theta\\ B_n=\frac{2}{\pi}\int_{0}^{\pi}h(\theta)\sin{2n\theta}d\theta$$ where $h(\theta):=g\left(2\cos(\theta),2\sin(\theta)\right)$