Closed form of a series

Question

Is there exist a closed form for the series of the form

$$ \sum_{k=0}^{[n/2]}(-a)^{k}\binom{n-k}{k} $$

where $0<a\leq1$. For example, we have $$ \sum_{k=0}^{[n/2]}\left(-\frac{1}{4}\right)^{k}\binom{n-k}{k}=\frac{n+1}{2^n}. $$

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Clearly, we can choose $\ds{b > 1}$ such that $\ds{\verts{a \over z\pars{z + 1}} < 1}$ when $\ds{\verts{z} = b}$. Then,

\begin{align}&\color{#66f}{\large \sum_{k\ =\ 0}^{\floor{n/2}}\pars{-a}^{k}{n - k \choose k}} =\sum_{k\ =\ 0}^{\infty}\pars{-a}^{k} \oint_{\verts{z}\ =\ b} {\pars{1 + z}^{n - k} \over z^{k + 1}}\,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ b}{\pars{1 + z}^{n} \over z}\sum_{k\ =\ 0}^{\infty} \bracks{-\,{a \over z\pars{1 + z}}}^{k}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}}{\pars{1 + z}^{n} \over z} {1 \over 1 + a/\bracks{z\pars{1 + z}}}\,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}}{\pars{1 + z}^{n + 1} \over z^{2} + z + a} \,{\dd z \over 2\pi\ic} =\oint_{\verts{z}}{\pars{1 + z}^{n + 1} \over \pars{z - z_{-}}\pars{z - z_{+}}}\, \,{\dd z \over 2\pi\ic}\,,\qquad z_{\pm} = {- 1 \pm \root{1 - 4a} \over 2} \end{align}

\begin{align}&\color{#66f}{\large \sum_{k\ =\ 0}^{\floor{n/2}}\pars{-a}^{k}{n - k \choose k}} =\oint_{\verts{z}}\pars{1 + z}^{n + 1} \pars{{1 \over z - z_{+}} - {1 \over z - z_{-}}}\,{1 \over z_{+} - z_{-}} \,{\dd z \over 2\pi\ic} \\[5mm]&={1 \over \root{1 - 4a}} \bracks{\pars{1 + z_{+}}^{n + 1} - \pars{1 + z_{-}}^{n + 1}} \\[5mm]&=\color{#66f}{\large{1 \over \root{1 - 4a}}\bracks{ \pars{1 + \root{1 - 4a} \over 2}^{n + 1}- \pars{1 - \root{1 - 4a} \over 2}^{n + 1}}} \end{align}

Answer 2

There is a general formula (a CAS found it) $$\sum_{k=0}^{n/2}(-a)^{k}\binom{n-k}{k}=\frac{ (1+A)^{n+1}-(1-A)^{n+1}}{2^{n+1}A}$$ where $A=\sqrt{1-4a}$.

To get the formula for $a=\frac 14$, you will need to consider the limit.