2

I need to find sum like $$\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1},\qquad \text{ for even } n$$

Example:

Find the sum of $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$

Martin Sleziak
  • 53,687
Sayan Ghosh
  • 99
  • interestingly enough for $n$ odd by symmetry you get this to be exactly half of the entire row, i.e. $2^{n-1}$ – gt6989b Jun 08 '16 at 20:34
  • Very similar question: http://math.stackexchange.com/questions/42797/evaluate-binomn0-binomn2-binomn4-cdots-binomn2k-cdots If you search a bit, you can probably find several other question on the same topic. – Martin Sleziak Jun 09 '16 at 10:12

4 Answers4

8

Hint: Consider the difference between: $$ \sum_{k=0}^{20}\binom{20}{k}1^k = (1+1)^{20}\quad\text{and}\quad \sum_{k=0}^{20}\binom{20}{k}(-1)^k = (1-1)^{20}.$$

Jack D'Aurizio
  • 353,855
4

The important binomial theorem states that

\begin{equation} \sum_{k=0}^n\binom{n}{k} x^ky^{n-k}=(x+y)^n \end{equation}

Setting $x=1$ and $y=1$, we have the following relation

\begin{equation} \sum_{k=0}^n\binom{n}{k} =2^n \end{equation}

and setting $x=-1$ and $y=1$, we have the following relation

\begin{equation} \sum_{k=0}^n\binom{n}{k} (-1)^k=0 \end{equation}

Hence we have

\begin{equation} \sum_{k=1}^{n/2}\binom{n}{2k-1} =\frac{1}{2}\left(\sum_{k=0}^n\binom{n}{k}-\sum_{k=0}^{n}\binom{n}{k}(-1)^k\right)=2^{n-1}. \end{equation}

This may be easier to see with your example

\begin{align} &\binom{20}{1}+\binom{20}{3}+\cdots+\binom{20}{19}=\\ &\frac{1}{2}\left[\color{red}{\binom{20}{0}}+\binom{20}{1}+\color{red}{\binom{20}{2}}+\cdots+\binom{20}{20}-\left(\color{red}{\binom{20}{0}}-\binom{20}{1}+\color{red}{\binom{20}{2}}+\cdots+\binom{20}{20}\right)\right] \end{align}

Sophie Agnesi
  • 2,377
2

If you know your Pascal's triangle, you know that $$\binom{20}{1} + \binom{20}{3} +..+ \binom{20}{19}= \sum_{k=0}^{19} \binom{19}{k}$$

Joffan
  • 39,627
2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} \color{#f00}{\sum_{k = 0}^{9}{20 \choose 2k + 1}} & = \sum_{k = 0}^{20}{20 \choose k}\half\bracks{1 - \pars{-1}^{k}} = \half\braces{\pars{1 + 1}^{20} - \bracks{1 + \pars{-1}}^{20}} \\[3mm] & = \color{#f00}{2^{19}} = 524288 \end{align}

Felix Marin
  • 89,464