Sum reminiscent of $(1+x)^N$ (binomial theorem)

Question

I stumbled upon this sum while working on my thesis:

$$\sum_{k=0}^N \binom{N}{2k} x^k$$

I know that

$$\sum_{k=0}^N \binom{N}{k} x^k = (1+x)^N$$

But when it comes to the sum above I'm lost. Is there any way to solve it?

PS $|x|<1$ and N is an even number. Also, as Semiclassical remarked, the sum actually runs up to $N/2$:

$$\sum_{k=0}^{N/2} \binom{N}{2k} x^k$$

Answer 1

Given the comment about the sum not even being defined for $2k>N$, I think that you could legitimately rewrite your question as $$\sum_{k=0}^N \binom{N}{k} {\sqrt{x}}^k$$ with the sum taken over even values of $k$ only.

In that case, observe that $$\sum_{k=0}^N \binom{N}{k} {\sqrt{x}}^k = (1+{\sqrt{x}})^N$$ (which you know) is the sum taken over even values of $k$ plus an unwanted sum taken over odd values of $k$, and $$\sum_{k=0}^N \binom{N}{k}(-{\sqrt{x}})^k = (1-{\sqrt{x}})^N$$ is the sum taken over even values of $k$ minus the same unwanted sum taken over odd values of $k$.

I hope you don't mind if I leave you to combine the two to get the answer you need.

Generalisation

It doesn't have to be $\binom{N}{2k}$ in your original question.

Consider $\binom{N}{3k}$ instead. Take $\omega$ to be a primitive cube root of 1: that is, one whose $n$th power is 1 if and only if $n$ is divisible by 3. (To be explicit, $\omega=\frac12({-1±\sqrt{3}})$). Take the sum $$\sum_{i=1}^3\sum_{k=0}^N \binom{N}{k} (\omega^i{\sqrt[3]{x}})^k$$ and note:

• This is the sum of three series each of whose sums you know.
• Leaving aside the binomial, the coefficient of $\sqrt[3]{x}^k$ is 3 when $k$ is divisible by 3 (because all the $\omega^k$ equal 1), and 0 when it isn't (because all the $\omega^k$ add up to 0).

So now you know how to get $$\sum_{k=0}^N \binom{N}{k} {\sqrt[3]{x}}^k$$ with the sum taken over values of $k$ divisible by 3.

The same thing works for all other multiples, but I thought it would be more readable to use "3" rather than yet another variable.