# Sum of combinatorics sequence $\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1}$

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 $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$

• interestingly enough for $n$ odd by symmetry you get this to be exactly half of the entire row, i.e. $2^{n-1}$ Commented Jun 8, 2016 at 20:34
• Very similar question: math.stackexchange.com/questions/42797/… If you search a bit, you can probably find several other question on the same topic. Commented Jun 9, 2016 at 10:12

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

The important binomial theorem states that

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

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

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

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

$$\sum_{k=0}^n\binom{n}{k} (-1)^k=0$$

Hence we have

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

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}

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

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\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}