Proving $\binom{2n}{n}<4^{n-1}$ for all positive integers $n\geq 5$ For $n=5$, we get $\binom{10}{5}<4^4$. Asumme inductively that $\binom{2n}{n}<4^{n-1}$ for all positive integers $n\geq 5$. Then I need to prove that $\binom{2(n+1)}{n+1}<4^{n}$ for all positive integers $n\geq 5$. I've been looking around a bit, and I've found that the expression $\binom{2n}{n}$ is precisely a central binomial coefficient. Thus $\sum_{k=0}^{n+1}\binom{n+1}{k}^2=\binom{2(n+1)}{n+1}$. How can I use this to finish my proof?

Using inductive argument, you can do something simpler.
$$\binom{2n+2}{n+1}=\frac{2(2n+1)}{n+1}\binom{2n}{n}$$
Now prove that
$$\frac{2(2n+1)}{n+1}<4.$$
Once the inequality is proven, we can multiply both sides by $\binom{2n}{n}$ and get,
$$\frac{2(2n+1)}{n+1}\binom{2n}{n}<4\binom{2n}{n}\to \binom{2n+2}{n+1}<4\binom{2n}{n}$$
and using the inductive hypothesis we get,
$$\binom{2n+2}{n+1}<4\binom{2n}{n}<4\cdot4^{n-1}=4^n.$$
