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How can I prove that $$2^n=2\left({n \choose 0}+{n \choose 2}+{n \choose 4}+\dots\right)$$ using the binomial theorem. I've tried expanding $(x-y)^n$ with multiple different values of $x$ and $y$ but nothing pans out. I can see why the above is true when looking at Pascal's triangle as the sum of the $n-1$ row gives $2^{n-1}$ and is indeed the sum above, but I can't seem to pin it down algebraically. Thank you.

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    $\begingroup$ Calculate $(1+x)^n+(1-x)^n$ using the Binomial Theorem, and then set $x=1$. $\endgroup$ – André Nicolas Oct 14 '14 at 17:15
  • $\begingroup$ I think you can use Pascal's rule to get it to something that looks like this. $\endgroup$ – Akiva Weinberger Oct 14 '14 at 17:18
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Writing $2^n$ as $$2^n=2^n+0^n=(1+1)^n+(-1+1)^n$$ and by applying the Binomial Theorem, you have that $$\begin{align*}2^n&=(1+1)^n+(-1+1)^n=\sum_{k=0}^{n}\binom{n}{k}1^k\cdot1^{n-k}+\sum_{k=0}^{n}\binom{n}{k}(-1)^k\cdot1^{n-k}=\\&=\sum_{k=0}^{n}\binom{n}{k}+\sum_{k=0}^{n}\binom{n}{k}(-1)^k\end{align*}$$ and since $$(-1)^k=\begin{cases}1, &\text{if } k \in \mathbb{2N}\\-1, &\text{if } k \in \mathbb{2N+1}\end{cases}$$ the above equation can be written as $$\begin{align*}2^n&=\sum_{k=0}^{n}\binom{n}{k}+\sum_{k \in \mathbb{2N}}^{n}\binom{n}{k}-\sum_{k \in \mathbb{2N+1}}^{n}\binom{n}{k}=2\sum_{k \in \mathbb{2N}}^{n}\binom{n}{k}=\\&\\&=2\left(\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\ldots\right)\end{align*}$$

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Hint: $$ (x+y)^n = \sum_{k=0}^n \binom nk x^ky^{n-k} $$Now change $y\to -y$ and make the sum and difference of both relations.

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take a set $S$ of $n$ things. you are computing the number of subsets even cardinality. it is well-known (or easily proved) that there are $2^n$ subsets in all, and that as many have even cardinality as have odd cardinality.

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