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?

Hint: Prove directly that
$
\dfrac{\binom{2(n+1)}{n+1}}{\binom{2n}{n}}<4
$
and conclude that
$
\binom{2(n+1)}{n+1} < 4 \binom{2n}{n}
$.
