Binomial summation with alternating terms?

Question

How do I solve problems of type :$$\sum_{k=1}^{(n+1)/2}\binom n{2k-1}x^k\text{ or }\sum_{k=0}^{n/2}\binom n{2k}x^k$$

I tried transferring the binomial to $n-1$ but the repeating $x^k$ makes it weird.

Edit: I started with $$(1+x)^n+(1-x)^n$$ with $x=\sqrt5$

Answer 1

Hint

Consider $$(1+x)^n+(1-x)^n=\sum_{k=0}^n\binom{n}{k}(1+(-1)^k)x^k$$ and $$ (1+x)^n-(1-x)^n=\sum_{k=0}^n\binom{n}{k}(1-(-1)^k)x^k $$ What can you say about $1+(-1)^k$ and $1-(-1)^k$ when $k$ is even or odd?

Answer 2

You can use the binomial theorem twice, $$ (1+x)^n=\sum_{k=0}^n{\binom{n}{k}x^k}\\ (1-x)^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}x^k} $$ If $E$ is the sum of the even terms, and $O$ the sum of the odd terms, we have $$ (1+x)^n=E+O\\ (1-x)^n=E-O $$

Answer 3

Okay, now that you've added some context, I think I can answer this question. The Binomial Theorem says $$ \left(1+\sqrt{x}\right)^n=\sum_{k=0}^n\binom{n}{k}x^{k/2}\tag1 $$ and $$ \left(1-\sqrt{x}\right)^n=\sum_{k=0}^n\binom{n}{k}(-1)^kx^{k/2}\tag2 $$ Adding $(2)$ to $(1)$ and dividing by $2$ gives $$ \begin{align} \frac12\left(\left(1+\sqrt{x}\right)^n+\left(1-\sqrt{x}\right)^n\right) &=\sum_{k=0}^n\binom{n}{k}\overbrace{\frac{1+(-1)^k}2}^{\text{$1$ when $k$ is even}}x^{k/2}\\ &=\sum_{k=0}^{n/2}\binom{n}{2k}x^k\tag3 \end{align} $$ Subtracting $(2)$ from $(1)$ and dividing by $2$ gives $$ \begin{align} \frac12\left(\left(1+\sqrt{x}\right)^n-\left(1-\sqrt{x}\right)^n\right) &=\sum_{k=0}^n\binom{n}{k}\overbrace{\frac{1-(-1)^k}2}^{\text{$1$ when $k$ is odd}}x^{k/2}\\ &=\sum_{k=1}^{(n+1)/2}\binom{n}{2k-1}x^{k-\frac12}\tag4 \end{align} $$