What's $\sum_{k=0}^n\binom{n}{2k}$?

Question

How do you calculate $\displaystyle \sum_{k=0}^n\binom{n}{2k}$? And doesn't the sum terminate when 2k exceeds n, so the upper bound should be less than n?

EDIT: I don't understand the negging. Have I violated a rule of conduct or something?

Answer 1

First of all, $\dbinom{n}{m} = 0$ if $m > n$. Hence, $\displaystyle\sum_{k = 0}^{n}\dbinom{n}{2k} = \sum_{k = 0}^{\lfloor n/2 \rfloor}\dbinom{n}{2k} = \sum_{\substack{0 \le m \le n \\ m \ \text{is even}}}\dbinom{n}{m}$.

To help you calculate the sum, note that by the binomial theorem:

$2^n = (1+1)^n = \displaystyle\sum_{m = 0}^{n}\dbinom{n}{m}1^m = \sum_{\substack{0 \le m \le n \\ m \ \text{is even}}}\dbinom{n}{m} + \sum_{\substack{0 \le m \le n \\ m \ \text{is odd}}}\dbinom{n}{m}$

$0 = (1-1)^n = \displaystyle\sum_{m = 0}^{n}\dbinom{n}{m}(-1)^m = \sum_{\substack{0 \le m \le n \\ m \ \text{is even}}}\dbinom{n}{m} - \sum_{\substack{0 \le m \le n \\ m \ \text{is odd}}}\dbinom{n}{m}$.

Do you see how to finish?

Answer 2

Recall that each binomial coefficient is the sum of the two above it in Pascal's triangle.

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\sum_{k = 0}^{n}{n \choose 2k}:\ {\large ?}}$

$$ \mbox{Note that}\quad\sum_{k = 0}^{n}{n \choose 2k} =\sum_{k = 0}^{\color{#c00000}{\large\infty}}{n \choose 2k}. $$

$$ \mbox{We'll use the identity}\quad {m \choose s}=\oint_{0\ <\ \verts{z}\ =\ a} {\pars{1 + z}^{m} \over z^{s + 1}}\,{\dd z \over 2\pi\ic}\,,\quad s \in {\mathbb N} $$

Then, \begin{align} &\color{#66f}{\large\sum_{k = 0}^{n}{n \choose 2k}} =\sum_{k = 0}^{\infty}\oint_{\verts{z}\ =\ a\ >\ 1} {\pars{1 + z}^{n} \over z^{2k + 1}}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ a\ >\ 1}{\pars{1 + z}^{n} \over z} \sum_{k = 0}^{\infty}\pars{1 \over z^{2}}^{k}\,{\dd z \over 2\pi\ic} \\[3mm]&=\oint_{\verts{z}\ =\ a\ >\ 1}{\pars{1 + z}^{n} \over z} {1 \over 1 - 1/z^{2}}\,{\dd z \over 2\pi\ic} =\oint_{\verts{z}\ =\ a\ >\ 1}{\pars{1 + z}^{n}\,z \over \pars{z - 1}\pars{z + 1}} \,{\dd z \over 2\pi\ic} \\[3mm]&={\pars{1 + 1}^{n}\times 1 \over 1 + 1}=\color{#66f}{\large 2^{n - 1}} \end{align}

Answer 4

$$\sum_{k=1}^n \binom{n}{2k}=2^{n-1}-1$$

BECAUSE:

$$\sum_{k=1}^n \binom{n}{2k}=\binom{n}{2}+\binom{n}{4}+\binom{n}{3}+ \dots$$

$$(x+y)^n=\sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$$

$$x=1,y=1: \sum_{k=0}^n \binom{n}{k}=\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+ \dots=2^n$$

$$\sum_{k=1}^n \binom{n}{2k}=\sum_{k=0}^n \binom{n}{2k}-1=\frac{\sum_{k=0}^n \binom{n}{k}}{2}-1=2^{n-1}-1$$