Middle elements of Pascal's Triangle

Since $\frac{1}{\sqrt{1-4x}} \; = \; \sum_{n=0}^\infty {2n \choose n}x^n \hspace{2cm} |x| < \tfrac14$ we deduce that $\sum_{n=0}^\infty f(n)x^n \; = \; \left(\sum_{n=0}^\infty {2n \choose n}x^n\right)^2 \; = \; \frac{1}{1-4x} \hspace{2cm} |x| < \tfrac14$ and hence it follows that $f(n) = 4^n$ for all $n \ge 0$ . Since $4 \equiv -1 \pmod{5}$ it is clear that $f(2017) = 4^{2017} \equiv \boxed{4} \pmod{5}$ .

$\displaystyle \text{Claim:} \binom{2r}{r} = {(-4)}^r \binom{-1/2}{r}$

The proof is left as an exercise to the reader but it is advised to learn it after you have proved it.

$\displaystyle \binom{2k}{k} = {(-4)}^k \binom{-1/2}{k}$

$\displaystyle \binom{2n-2k}{n-k} = {(-4)}^{n-k} \binom{-1/2}{n-k}$

$\displaystyle \binom{2k}{k} \binom{2n-2k}{n-k} = {(-4)}^n \binom{-1/2}{k} \binom{-1/2}{n-k}$

$\displaystyle \sum_{k=0}^n{\binom{2k}{k} \binom{2n-2k}{n-k}} = \sum_{k=0}^n{{(-4)}^n \binom{-1/2}{k} \binom{-1/2}{n-k}}$

Now we can use Vandermonde's convolution,

$\displaystyle = {(-4)}^n \sum_{k=0}^n{\binom{-1/2}{k}\binom{-1/2}{n-k}} = {(-4)}^n \binom{-1}{n} = 4^n$