Prove that $2^{n-1}$ divides $\binom{n}{1} + \binom{n}{3}5 + \binom{n}{5}25 + \binom{n}{7}125 + \cdots$ for $n \geqslant 1$. Prove that $2^{n-1}$ divides $\binom{n}{1} + \binom{n}{3}5 + \binom{n}{5}25 + \binom{n}{7}125 + \cdots$ for $n \geqslant 1$.
Assume $\binom{n}{k} = 0$ if $k>n$.
Does anyone know an elementary proof without using the Binet formula?
 A: Here's a combinatorial-based argument.  
Question: How many ways are there to tile a $1 \times (n-1)$ board with black and white squares (which take up one space) and red, yellow, green, and violet dominoes (which take up two spaces)?  
Answer 1: $2^{n-1} F_n$. 
Proof of Answer 1: Since each of the four dominoes can be mapped to a coloring of the original spaces as black or white (Red = BB, Yellow = WW, Green = BW, Violet = WB), the answer is the same as the number of ways to color each of the $n-1$ spaces in a $1 \times (n-1)$ board black or white and then tile the board using squares and dominoes.  There are $2^{n-1}$ ways to color the $n-1$ spaces, and then $F_n$ ways to tile the board with squares and dominoes.  So the answer is $2^{n-1} F_n$.  (Here we are using the standard combinatorial interpretation of the Fibonacci numbers $F_n$.)  
Answer 2: $\sum_{k \geq 0} \binom{n}{2k+1} 5^k$. 
Proof of Answer 2: A $1 \times (n-1)$ board has $n$ grooves between spaces (including the groove before the first space and the groove after the last space).  Out of those $n$ grooves, choose an odd number $2k+1$ of them: $a_0, a_1, a_2, \ldots, a_{2k}$.  Color black the spaces between grooves $0$ and $a_0$, and for each $1 \leq i \leq k$, the spaces between grooves $a_{2i-1}$ and $a_{2i}$.  Color white the spaces between grooves $a_{2i-2}$ and $a_{2i-1}$, $1 \leq i \leq k$, as well as the spaces between grooves $a_{2k}$ and $n$.  For example, if $n = 11$ and we choose grooves $1, 3, 4, 5, 9$, the coloring will look like BWWBWBBBBW.  The result is a coloring of the board with black and white tiles.  By conditioning on the number of transitions from white to black, this gives a combinatorial proof of the well-known formula $\sum_{k \geq 0} \binom{n}{2k+1} = 2^{n-1}$.
Now, to create a tiling of a $1 \times (n-1)$ board with black and white squares and our four colored dominoes, first create a tiling of the board with black and white squares only, as described above.  Then replace one or more of the WB pairs that represents a transition from white spaces to black spaces with a colored domino.  Any resulting tiling can be easily mapped back to a tiling with black and white squares only by replacing each colored domino with a WB pair.  For example, the tiling RRRRWWWWBW (with RR representing a red domino) maps back to WBWBWWWWBW.
If there are $k$ WB transitions in a tiling, then there are $\binom{n}{2k+1}$ ways to choose the required $2k+1$ grooves from $n$, and $5^k$ choices for what to do with each WB pair (either keep the WB pair or replace it with one of four colored dominoes).  Multiplying together and summing over all values of $k$ gives Answer 2.
Conclusion:
Our two answers produce $$\sum_{k \geq 0} \binom{n}{2k+1} 5^k = 2^{n-1} F_n.$$
Since the Fibonacci numbers are all integers, $\sum_{k \geq 0} \binom{n}{2k+1} 5^k$ is divisible by $2^{n-1}$.
(This argument is a modified and simplified version of a combinatorial proof of a similar identity in Proofs That Really Count, p. 130.)
A: This proof may not qualify as elementary, as it uses some algebraic number theory.
Let $R=\mathbb{Z}[(1+\sqrt{5})/2]$, the ring of integers of $\mathbb{Q}(\sqrt{5})$, and define $\alpha:=(1+\sqrt{5})/2$ and $\beta:=(1-\sqrt{5})/2\in R$. The binomial theorem implies
$$
\begin{align*}
S_n:=
{n\choose 1}+5{n\choose 3}+25{n\choose 5}\ldots&=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2\sqrt{5}}\\
&=2^{n-1}\frac{\alpha^n-\beta^{n}}{\sqrt{5}}.
\end{align*}
$$
Now $2$ and $\sqrt{5}$ are coprime, so $2^{n-1}$ must divide $S_n$ in $R$. But the only rational numbers in $R$ are integers, so $2^{n-1}$ divides $S_n$ as integers.
A: Note that $\binom{n-2}{j-1}+2\binom{n-2}{j}+\binom{n-2}{j+1}=\binom{n}{j+1}$.  Define
$$
g(n)=\sum_j \binom{n}{2j+1}\cdot5^j.
$$
Then
$$
\begin{aligned}
4g(n-2)+2g(n-1)&=4\sum_j\binom{n-2}{2j+1}\cdot5^j+2\sum_j\binom{n-1}{2j+1}\cdot5^j\\
&=(5-1)\sum_j\binom{n-2}{2j+1}\cdot5^j+2\sum_j\left[\binom{n-2}{2j}+\binom{n-2}{2j+1}\right]\cdot5^j\\
&=\sum_j\binom{n-2}{2j+1}\cdot5^{j+1}+2\sum_j\binom{n-2}{2j}\cdot5^j+\sum_j\binom{n-2}{2j+1}\cdot5^j\\
&=\sum_j\left[\binom{n-2}{2j-1}+2\binom{n-2}{2j}+\binom{n-2}{2j+1}\right]\cdot5^j\\
&=\sum_j\binom{n}{2j+1}\cdot5^j\\
&=g(n).
\end{aligned}
$$
So $g(n)=2g(n-1)+4g(n-2)=2\left[g(n-2)+2g(n-2)\right]$ with the initial conditions $g(1)=1$, $g(2)=2$.  The result follows by induction.
It would be satisfying to interpret this calculation combinatorially.  I'll think about this.
