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*} A: (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
Hint:
We know from the binomial theorem that, $$ (1+x)^{1/2}=\sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k $$ (for suitable $x$).
This means, in particular, that $$ 1+x=(1+x)^{1/2}\cdot(1+x)^{1/2}=\left[\sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k\right]^2. $$ Note that for any $n\geq 2$, the coefficient by $x^n$ on the left side is $0$. Therefore the coefficient by $x^n$ (where $n\geq2$) must also be $0$ on the right side.
The coefficient by $x^n$ in the right side can be computed using the Cauchy product.