Use Newton's Binomial Theorem to prove series Verify that for all integer $n\ge 2$,
\begin{eqnarray*}
\sum_{k=0}^n\binom{\frac{1}{2}}{k}\binom{\frac{1}{2}}{n-k}=0.
\end{eqnarray*}
(Hint: Consider Newton's Binomial Theorem.)
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EDIT:
From Newton's Binomial Theorem:
\begin{eqnarray*}
(1+x)^p &=& \sum_{n=0}^\infty \begin{pmatrix}
p \\n
\end{pmatrix} x^n.
\end{eqnarray*}
So,
\begin{eqnarray*}
(1+x)^\frac{1}{2} &=& \sum_{k=0}^\infty \begin{pmatrix}
\frac{1}{2} \\k
\end{pmatrix} x^k\\
B: (1+x)^\frac{1}{2} &=& \sum_{k=n}^\infty \begin{pmatrix}
\frac{1}{2} \\{n-k}
\end{pmatrix} x^{n-k}
\end{eqnarray*}
I assume finding $A-B$ will lead to the answer $0$ but I'm not sure how to continue from here
