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I'll be happy if you could help me prove this argument with algebraic tools:

$${N\choose 0}a^N+{N\choose 1}a^{N-2}+{N\choose 2}a^{N-4}+{N\choose 3}a^{N-6}+\dots = \frac{a^2+1}{a}$$

Thanks, Don

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oops, editing is out of sync. Sorry @Zev – user13838 Oct 11 '11 at 20:20
Very out of sync, I hope I didn't break too much editing. $\LaTeX>>$ image. – Asaf Karagila Oct 11 '11 at 20:21
This question has gone through 6 edits within 8 minutes. Perhaps we should let the post settle down a bit :) – Srivatsan Oct 11 '11 at 20:24
I don't think the formula is correct. I get $(a+1/a)^n$ for the binomial sum. – Mike Spivey Oct 11 '11 at 20:25
That's really an equality, rather than an "argument". – Arturo Magidin Oct 11 '11 at 20:29

The summation terminates at $\binom{N}{N}$, because once you get to $\binom{N}{k}$ with $k\gt N$, the binomial coefficient is zero. So the left hand side is equal to $$\begin{align*} \binom{N}{0}a^N &+ \binom{N}{1}a^{N-2} + \cdots + \binom{N}{N}a^{N-2N}\\ &\qquad= \binom{N}{0}a^N\left(\frac{1}{a}\right)^0 + \binom{N}{1}a^{N-1}\left(\frac{1}{a}\right)^1 + \cdots + \binom{N}{N}a^0\left(\frac{1}{a}\right)^N.\end{align*}$$ This is equal to $$\left(a+\frac{1}{a}\right)^N = \left(\frac{a^2+1}{a}\right)^N$$ by the Binomial Theorem.

So you are missing an exponent $N$ on the right.

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