- #1
Quesadilla
- 95
- 13
Homework Statement
From an old exam: Show that
\begin{equation*}
\sum_{0 \leq 2k \leq n} \binom{n}{2k}2^k = 0 (3) \text{ iff } n = 2 (4).
\end{equation*}
By ##a = b (k)## I mean that ##a## is congruent to ##b## modulo ##k##.
Homework Equations
Binomial theorem: ## (a + b)^m = \sum_{k=1}^m \binom{k}{m} a^k b^{m-k}##.
The Attempt at a Solution
A hint was provided to consider ##\frac{1}{2}((1 + \sqrt{2})^n + (1 - \sqrt{2})^n) ##.
I have taken some baby steps. Using the binomial theorem, I was able to establish
\begin{equation*}
\sum_{0 \leq 2k \leq n} \binom{n}{2k}2^k = \frac{1}{2}((1 + \sqrt{2})^n + (1 - \sqrt{2})^n).
\end{equation*}
Now in other words I need to find for exactly which ##n## the above is divisible by ##3##, which I guess is equivalent to saying that ## ((1 + \sqrt{2})^n + (1 - \sqrt{2})^n) ## is divisible by ##6##.
I would greatly appreciate some small hints.