Binomial theorem and sum from $k = 0$ to $N/2$

Question

While calculating a partition function in physics, I stumbled across a weird sum, namely:

$$Z=\sum_{k=0}^{N/2}{N \choose 2k}b^{N-k}e^{ak}$$

Wolfram says that it's equal to:

$$Z = \frac{1}{2}\left((b-\sqrt{b}e^{a/2})^N+(b+\sqrt{b}e^{a/2})^N\right)=\frac{(x-y)^N+(x+y)^N}{2}$$

But I dont understand how we can link it to the binomial theorem since the coefficient is: ${N \choose 2k}$ and not ${N/2 \choose 2k}$ and then prove the formula.

Answer 1

It might be easier to see what is going on if we write out some terms: $$ \begin{align} &\overbrace{\sum_{k=0}^N\binom{N}{k}b^{N-k/2}e^{ak/2}}^{b^{N/2}\left(\sqrt{b}+e^{a/2}\right)^N}+\overbrace{\sum_{k=0}^N(-1)^k\binom{N}{k}b^{N-k/2}e^{ak/2}}^{b^{N/2}\left(\sqrt{b}-e^{a/2}\right)^N}\tag1\\[6pt] &=\phantom{2}\binom{N}{0}b^Ne^0+\binom{N}{1}b^{N-1/2}e^{a/2}+\phantom{2}\binom{N}{2}b^{N-1}e^a+\binom{N}{3}b^{N-3/2}e^{3a/2}+\cdots\tag{2a}\\ &+\phantom{2}\binom{N}{0}b^Ne^0-\binom{N}{1}b^{N-1/2}e^{a/2}+\phantom{2}\binom{N}{2}b^{N-1}e^a-\binom{N}{3}b^{N-3/2}e^{3a/2}+\cdots\tag{2b}\\[6pt] &=2\binom{N}{0}b^Ne^0\phantom{{}+\binom{N}{1}b^{N-1/2}e^{a/2}}+2\binom{N}{2}b^{N-1}e^a\phantom{{}+\binom{N}{3}b^{N-3/2}e^{3a/2}}+\cdots\tag3\\[6pt] &=2\sum_{j=0}^{\lfloor N/2\rfloor}\binom{N}{2j}b^{N-j}e^{aj}\tag4 \end{align} $$ The terms with even $k$ in $(1)$ become the terms with $2j$ in $(4)$ and the term with odd $k$ in $(1)$ cancel each other. Since $k$ goes from $0$ to $N$, $j$ goes from $0$ to $\lfloor N/2\rfloor$.

We can substitute $j\mapsto k$ in $(4)$, if desired.

Answer 2

Because $$\frac{1+(-1)^k}{2}=\begin{cases}1 &\text{if $k$ is even}\\0 &\text{if $k$ is odd}\end{cases}$$ we have $$\sum_{k \ge 0} c_{2k} = \sum_{k \ge 0} \frac{1+(-1)^k}{2} c_k.$$ Now take $c_k=\binom{N}{k}b^{N-k/2}e^{ak/2}$ and apply the binomial theorem to each of the two resulting sums.

Answer 3

$$Z=\sum_{k=0}^{N/2}{N \choose 2k}b^{N-k}e^{ak}=b^N\sum_{k=0}^{N/2}{N \choose 2k}\left(\sqrt{\frac{e^a}{b}}\right)^{2k}$$ $$\sum_{k=0}^{N/2}{N \choose 2k}x^{2k}=\frac{1}{2} \left(\left(1-x\right)^N+\left(1+x\right)^N\right)$$ Make $x=\sqrt{\frac{e^a}{b}}$ to get the formula