The question I am trying to solve involves calculating the voltage function $V(r, \theta)$ ($\theta$ is the zenith angle) inside a spherical shell with radius $R$, where the voltage along the spherical shell is defined as follows: $$ V(R, \theta) = \begin{cases} V_0 & \theta < \frac{\pi}{2} \\ -V_0 & \theta > \frac{\pi}{2} \end{cases} $$
And the voltage follows the relation: $\nabla^2 V=0$.
I arrived at the following equation which I can't simplify. Below that, I've explained the steps I took to arrive at this equation. Judging by the final solution to this problem, I am almost sure that the following equation should simplify quite nicely but I can't simplify this any further.
$$ \sum_{i=0}^{n-1} (-1)^i \begin{pmatrix}n\\n-i\end{pmatrix} \begin{pmatrix}n\\i+1\end{pmatrix} $$
The only clue I could think of was the following relation, but I couldn't find any helpful connections.
$$ \sum_{i=0}^{n-1} \begin{pmatrix}n\\n-i\end{pmatrix} \begin{pmatrix}n\\i+1\end{pmatrix} = \begin{pmatrix}2n \\ n + 1 \end{pmatrix} $$
Alternatively, I was wondering if there is an easier way to solve this question to avoid arriving at these equations.
Thanks!
EDIT: By pure luck, I guessed the solution. I still do not know how to prove it. When $n$ is an even number, the result is $0$. (As one could prove using symmetry) When $n$ is odd and $n=2l + 1$:
$$ \sum_{i=0}^{n-1} (-1)^i \begin{pmatrix}n\\n-i\end{pmatrix} \begin{pmatrix}n\\i+1\end{pmatrix} = \begin{pmatrix}2l + 1 \\ l \end{pmatrix} (-1)^l $$
Full solution:
After solving the Laplacian equation, and the assumption that $V$ is independent of the azimuthal angle $\phi$, I got the following relation:
$$ V(r, \theta) = \sum_{l=0}^{\infty} \alpha_l r^l P_l(\cos(\theta)) $$
Where $\alpha_l$ is a constant(which should be calculated with respect to the initial condition), and $P_n(.)$ is the Legendre polynomial of degree n, with the following properties:
$$ P_n(x) = \frac{1}{2^n n!} \dfrac{d^n}{dx^n} \left[(x^2 - 1)^n\right]$$
$$ \langle P_n(x), P_m(x) \rangle = \delta_{nm} \dfrac{2}{2n + 1} $$
By taking the inner product of both sides of the following equation with $P_n(x)$:
$$ \sum_{l=0}^{\infty} \alpha_l r^l P_l(\cos(\theta)) = \begin{cases} V_0 & \theta < \frac{\pi}{2} \\ -V_0 & \theta > \frac{\pi}{2} \end{cases} $$
I get the following relation:
$$ \dfrac{\alpha_n R^n 2^{n+1} n!}{V_0 (2n + 1)} = \int_{0}^{1} \dfrac{d^n}{dx^n} \left[ (x^2-1)^n \right] dx - \int_{-1}^{0} \dfrac{d^n}{dx^n} \left[ (x^2-1)^n \right] dx $$
The right hand side is $0$ when $n$ is an even number. (as the integrand becomes an even function)
However, if the equation is simplified with the assumption that $n$ is an odd number, we arrive at the following relation. (Notice $(x^2 - 1)^n = (x-1)^n (x+1)^n$)
$$ \dfrac{\alpha_n R^n 2^n n}{V_0 (2n + 1)} = \sum_{i=0}^{n-1}(-1)^i \begin{pmatrix}n \\ n-i\end{pmatrix} \begin{pmatrix} n \\ i+1 \end{pmatrix}$$
If the right hand side is simplified, all is required to do is to solve for $\alpha_n$ and substitute the result for the general solution found at the beginning of this solution.
After substituting the solution I guessed I got the following:
$$ V(r, \theta) = \sum_{l=0}^{\infty} (-1)^l V_0 \dfrac{r^{2l + 1}}{R^{2l + 1}} \dfrac{(2l)!}{(2^l l!)^2} \times \dfrac{4l + 3}{2l + 2} P_l (\cos(\theta)) $$
Since there is no charge inside the sphere, the potential must vanish at the origin so \begin{align*} V(r, \theta) = \sum_{\ell=0}^\infty A_\ell r^\ell P_\ell(\cos(\theta)) \end{align*}
By orthogonality, we can compute the $A_\ell$ with the following equation (this holds in the azimuthal symmetry case only): \begin{align*} A_\ell &= \frac{2\ell + 1}{2}\frac{1}{R^\ell} \int_0^\pi \sin(\theta) V(R,\theta)P_\ell(\cos(\theta))d\theta\\ &= \frac{2\ell + 1}{2}\frac{V_0}{R^\ell} \left(\int_0^{\pi/2}\sin(\theta)P_\ell(\cos(\theta))d\theta - \int_{\pi/2}^\pi \sin(\theta)P_\ell(\cos(\theta))d\theta\right) \end{align*}
Now let $x = \cos(\theta)$ so the integrals become: \begin{align*} A_\ell &=\frac{2\ell + 1}{2}\frac{V_0}{R^\ell} \left(\int_0^1P_\ell(x)dx - \int_1^0 P_\ell(x)dx\right)\\ &= (2\ell + 1)\frac{V_0}{R^\ell}\int_0^1 P_\ell(x)dx \end{align*}
Since the $P_\ell$'s are even for even $\ell$ and odd for odd $\ell$, all the even terms go to $0$.
On the Wolfram Mathworld page for Legendre polynomials they present the formula \begin{align*} \int_0^1 P_\ell(x)dx = \frac{P_{\ell-1}(0) - P_{\ell+1}(0)}{2\ell + 1} \end{align*}
This is not a complete answer as I don't know how to proceed, but it is a another - though very similar - way to attempt the problem. Hopefully it helps and someone can find a way to complete this.