TITLE: Show that $ \sum_{k=0}^{n} \binom{2n+1}{2k} 3^k $ is divisible by $2^n$
QUESTION [8 upvotes]: I want to prove that
$$
 \sum_{k=0}^{n} \binom{2n+1}{2k} 3^k = \sum_{k=0}^{2n} \binom{2n}{k} 3^{\lceil k/2 \rceil}
$$
is divisible by $2^n$. 
I tried induction the following way
\begin{align*}
  \sum_{k=0}^{n} \binom{2n+1}{2k} 3^k
   & = \sum_{k=0}^{n} \left( \binom{2n}{2k-1} + \binom{2n}{2k} \right) 3^k \\
   & = \sum_{k=0}^{n} \left( \binom{2n-1}{2k-2} + \binom{2n-1}{2k-1} + \binom{2n-1}{2k-1} + \binom{2n-1}{2k} \right) 3^k \\
   & = 4\cdot \sum_{k=0}^{n-1} \binom{2(n-1)+1}{2k} 3^k + 2\sum_{k=0}^n \binom{2n-1}{2k-1} 3^k
\end{align*}
now I can apply the induction hypothesis to the first part, hence the entire first summand is divisible by $2^{n+1}$, if I can show that
$$
 \sum_{k=0}^n \binom{2n-1}{2k-1} 3^k
$$
is divisble by $2^{n-1}$ I would be done. But here I failed.

So do you have any hints how to show the original claim?

EDIT: As mentioned by Will Orrick in the comments, the inductive approach works too, just prove the stronger claim that
$$
 \sum_{k=0}^{n}\binom{2n}{2k}3^k, \quad
 \sum_{k=0}^{n}\binom{2n+1}{2k}3^k, \quad
 \sum_{k=0}^{n}\binom{2n+1}{2k+1}3^k, \quad
 \sum_{k=0}^{n}\binom{2n}{2k+1}3^k \quad
$$
are all divisible by $2^n$. If $n = 1$ we get $4$, $10$, $6$ and $2$ and if $n > 1$ we have
\begin{align*}
 & \sum_{k=0}^{n}\binom{2n}{2k}3^k \\
 & = 3 \sum_{k=0}^n \binom{2(n-1)+1}{2(k-1)+1} 3^{k-1}
     + \sum_{k=0}^n \binom{2(n-1)+1}{2k} 3^k \\
 & = 3 \sum_{k=0}^{n-1} \binom{2(n-1)+1}{2k+1} 3^k + 
       \sum_{k=0}^{n-1} \binom{2(n-1)+1}{2k} 3^k \\
 & = 3 \sum_{k=0}^{n-1} \binom{2(n-1)}{2k} 3^k + 3 \sum_{k=0}^{n-1} \binom{2(n-1)}{2k+1} 3^k
    + \sum_{k=0}^{n-1} \binom{2(n-1)}{2k-1} 3^k + \sum_{k=0}^{n-1} \binom{2(n-1)}{2k} 3^k \\
 & = 4 \sum_{k=0}^{n-1} \binom{2(n-1)}{2k} 3^k + 3 \sum_{k=0}^{n-2} \binom{2(n-1)}{2k+1} 3^k + \sum_{k=1}^{n-1} \binom{2(n-1)}{2k-1} 3^k \\
& = 4 \sum_{k=0}^{n-1} \binom{2(n-1)}{2k} 3^k + \sum_{k=1}^{n-1} \binom{2(n-1)}{2k-1} 3^k + \sum_{k=1}^{n-1} \binom{2(n-1)}{2k-1} 3^k \\
& = 4 \sum_{k=0}^{n-1} \binom{2(n-1)}{2k} 3^k + 6\sum_{k=0}^{n-2} \binom{2(n-1)}{2k+1} 3^k \\
& = 4 \sum_{k=0}^{n-1} \binom{2(n-1)}{2k} 3^k + 6\sum_{k=0}^{n-1} \binom{2(n-1)}{2k+1} 3^k  
\end{align*}
and now by assumption both sums are divisible by $2^{n-1}$, and as they are multiplied with even numbers their sum is divisible by $2^n$. Similar
\begin{align*}
 \sum_{k=0}^{n}\binom{2n+1}{2k}3^k
 & = \sum_{k=0}^n \left( \binom{2n-1}{2k-2} + \binom{2n-1}{2k-1} + \binom{2n-1}{2k-1} + \binom{2n-1}{2k } \right) 3^k \\
 & = \sum_{k=1}^n \binom{2n-1}{2k-2} 3^k + \sum_{k=0}^{n-1} \binom{2n-1}{2k} 3^k + 2\sum_{k=1}^n \binom{2n-1}{2k-1} 3^k \\
 & = 3 \sum_{k=0}^{n-1} \binom{2n-1}{2k} 3^k + \sum_{k=0}^{n-1} \binom{2n-1}{2k} 3^k + 2\sum_{k=1}^n \binom{2n-1}{2k-1} 3^k \\
& = 4 \sum_{k=0}^{n-1} \binom{2n-1}{2k} 3^k +  2\sum_{k=1}^n \binom{2n-1}{2k-1} 3^k \\
& = 4 \sum_{k=0}^{n-1} \binom{2(n-1)+1}{2k} 3^k +  6\sum_{k=0}^{n-1} \binom{2(n-1)+1}{2k+1} 3^k \\
\end{align*}
and for the other sums we have
\begin{align*}
 \sum_{k=0}^{n}\binom{2n+1}{2k+1}3^k
 & = 4 \sum_{k=0}^{n-1} \binom{2(n-1)+1}{2k+1} 3^k +  2\sum_{k=0}^n \binom{2n-1}{2k} 3^k\\
 & = 4 \sum_{k=0}^{n-1} \binom{2(n-1)+1}{2k+1} 3^k +  2\sum_{k=0}^{n-1} \binom{2(n-1)+1}{2k} 3^k\\
 \sum_{k=0}^{n}\binom{2n}{2k+1}3^k
 & = 4 \sum_{k=0}^{n-1} \binom{2(n-1)}{2k+1} 3^k +  2\sum_{k=0}^{n-1} \binom{2(n-1)}{2k} 3^k
\end{align*}
and the induction hypothesis gives the result.

REPLY [2 votes]: We show the following binomial identity for integral $n\geq 0$ is valid:
  \begin{align*}
\sum_{k=0}^n\binom{2n+1}{k}3^k=\color{blue}{2^n}\left(\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}3^k2^{n-2k}
+\sum_{k=0}^{\lfloor n/2  \rfloor-1}\binom{n}{2k+1}3^{k+1}2^{n-(2k+1)}\right)\tag{1}
\end{align*}
  We have a factor $\color{blue}{2^n}$ at the right-hand side of (1). Since the sums in parentheses are non-negative integers the claim follows immediately.

We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{\binom{2n+1}{2k}3^k}\\
&=\sum_{k=0}^{2n+1}\binom{2n+1}{k}(\sqrt{3})^k\frac{1+(-1)^k}{2}\\
&=\frac{1}{2}\sum_{k=0}^{2n+1}\binom{2n+1}{k}(\sqrt{3})^k+\frac{1}{2}\sum_{k=0}^{2n+1}\binom{2n+1}{k}(-\sqrt{3})^k\\
&=\frac{1}{2}\left(1+\sqrt{3}\right)^{2n+1}+\frac{1}{2}\left(1-\sqrt{3}\right)^{2n+1}\\
&=\frac{1}{2}\left(1+\sqrt{3}\right)\left(4+2\sqrt{3}\right)^{n}+\frac{1}{2}\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)^{n}\\
&=2^{n-1}\left(1+\sqrt{3}\right)\sum_{k=0}^n\binom{n}{k}(\sqrt{3})^k2^{n-k}
+2^{n-1}\left(1-\sqrt{3}\right)\sum_{k=0}^n\binom{n}{k}(-\sqrt{3})^k2^{n-k}\\
&=2^{n}\sum_{k=0}^n\binom{n}{k}(\sqrt{3})^k2^{n-k}\frac{1+(-1)^k}{2}
+2^{n}\sqrt{3}\sum_{k=0}^n\binom{n}{k}(-\sqrt{3})^k2^{n-k}\frac{1-(-1)^k}{2}\\
&=2^{n}\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}3^k2^{n-2k}
+2^{n}\sum_{k=0}^{\lfloor n/2 \rfloor -1}\binom{n}{2k+1}3^{k+1}2^{n-(2k+1)}\\
&\,\,\color{blue}{=2^n\left(\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}3^k2^{n-2k}
+\sum_{k=0}^{\lfloor n/2 \rfloor-1}\binom{n}{2k+1}3^{k+1}2^{n-(2k+1)}\right)}
\end{align*}
and the claim (1) follows.