HMMT February

First Solution: Note that $\binom{2 n}{n} =\frac{(2 n)!}{n!\cdot n!}=\frac{(2 n)(2 n-2)(2 n-4) \cdots(2)}{n!} \cdot \frac{(2 n-1)(2 n-3)(2 n-5) \cdots(1)}{n!} =2^{n} \cdot \frac{(-2)^{n}}{n!}(-\frac{1}{2})(-\frac{1}{2}-1)(-\frac{1}{2}-2) \cdots(-\frac{1}{2}-n+1) =(-4)^{n}\binom{-\frac{1}{2}}{n}$. Then, by the binomial theorem, for any real $x$ with $|x|<\frac{1}{4}$, we have $(1-4 x)^{-1 / 2}=\sum_{n=0}^{\infty}\binom{-\frac{1}{2}}{n}(-4 x)^{n}=\sum_{n=0}^{\infty}\binom{2 n}{n} x^{n}$. Therefore, $\sum_{n=0}^{\infty}\binom{2 n}{n}(\frac{1}{5})^{n}=\frac{1}{\sqrt{1-\frac{4}{5}}}=\sqrt{5}$. Second Solution: Consider the generating function $f(x)=\sum_{n=0}^{\infty}\binom{2 n}{n} x^{n}$. It has formal integral given by $g(x)=I(f(x))=\sum_{n=0}^{\infty} \frac{1}{n+1}\binom{2 n}{n} x^{n+1}=\sum_{n=0}^{\infty} C_{n} x^{n+1}=x \sum_{n=0}^{\infty} C_{n} x^{n}$ where $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$th Catalan number. Let $h(x)=\sum_{n=0}^{\infty} C_{n} x^{n}$; it suffices to compute this generating function. Note that $1+x h(x)^{2}=1+x \sum_{i, j \geq 0} C_{i} C_{j} x^{i+j}=1+x \sum_{k \geq 0}(\sum_{i=0}^{k} C_{i} C_{k-i}) x^{k}=1+\sum_{k \geq 0} C_{k+1} x^{k+1}=h(x)$ where we've used the recurrence relation for the Catalan numbers. We now solve for $h(x)$ with the quadratic equation to obtain $h(x)=\frac{1 / x \pm \sqrt{1 / x^{2}-4 / x}}{2}=\frac{1 \pm \sqrt{1-4 x}}{2 x}$. Note that we must choose the - sign in the $\pm$, since the + would lead to a leading term of $\frac{1}{x}$ for $h$ (by expanding $\sqrt{1-4 x}$ into a power series). Therefore, we see that $f(x)=D(g(x))=D(x h(x))=D(\frac{1-\sqrt{1-4 x}}{2})=\frac{1}{\sqrt{1-4 x}}$ and our answer is hence $f(1 / 5)=\sqrt{5}$.