Proof of identity involving binomial coefficients

Question

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

Answer 1

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.