Prove using induction that $2^n < \binom{2n}{n} < 4^n$ for $n \geq 2$

Question

Trying to prove that, for $n\geq2$, $2^n < \binom{2n}{n} < 4^n$.

Inductive hypothesis: Assume $P(k)$ is true: \begin{align} 2^k < \binom{2k}{k} < 4^n \\\\ \end{align}

Show $P(k+1)$

\begin{align} 2^{k+1} < \binom{2k+2}{k+1} < 4^{k+1} \\\\ \end{align}

Not sure what's the strategy here... since it's a compound inequality I think it make sense to break the two apart and prove them separately?

Start with the inductive hypothesis

\begin{align} 2^k < \binom{2k}{k} < 4^n \\\\ \end{align}

Trying to prove the first inequality... \begin{align} 2^k < \binom{2k}{k} \\\\ (2)2^k < (2)\binom{2k}{k} \\\\ 2^{k+1} < (2)\binom{2k}{k} \end{align}

Stuck here now... not sure if I see what to do to the equation to get to \begin{align} 2^{k+1} < \binom{2k+2}{k+1} \end{align}

Answer 1

For the induction step on the first inequality, you wish to show that if $$2^k < \binom{2k}{k}$$ then $$2^{k + 1} < \binom{2k + 2}{k + 1}$$ Using the definition of the binomial coefficient yields \begin{align*} \binom{2k + 2}{k + 1} & = \frac{(2k + 2)!}{(k + 1)!(k + 1)!}\\ & = \frac{(2k + 2)(2k + 1)(2k)!}{(k + 1)k!(k + 1)k!}\\ & = \frac{2(k + 1)(2k + 1)(2k)!}{(k + 1)k!(k + 1)k!}\\ & = 2 \cdot \frac{2k + 1}{k + 1} \cdot \frac{(2k)!}{k!k!}\\ & = 2 \cdot \frac{2k + 1}{k + 1}\binom{2k}{k}\\ & > 2\binom{2k}{k}\\ & > 2 \cdot 2^k & \text{by the induction hypothesis}\\ & = 2^{k + 1} \end{align*}

Answer 2

You can use the identity ${2k+2\choose k+1}={2k+1\choose k+1}+{2k+1\choose k}={2k\choose k+1}+2{2k\choose k}+{2k\choose k-1}=2{2k\choose k}+2{2k\choose k-1}>2{2k\choose k}$.

For the other side, $2{2k\choose k}+2{2k\choose k-1}\leq 4{2k\choose k}$