Sum with binomial coefficients: $\sum_{k=0}^{n}{2n\choose 2k}$ I'm repeating material for test and I came across the example that I can not do. How to calculate this sum:
$\displaystyle\sum_{k=0}^{n}{2n\choose 2k}$?
 A: $$(1+1)^{2n}= \displaystyle\sum_{k=0}^{2n}{2n\choose k}$$
$$(1-1)^{2n}= \displaystyle\sum_{k=0}^{2n}(-1)^k{2n\choose k}$$
Add them together.
OR Second solution:
You can use the formula
$${2n\choose 2k}={2n-1\choose 2k}+{2n-1\choose 2k-1}$$ to prove that
$$\displaystyle\sum_{k=0}^{n}{2n\choose 2k}=\displaystyle\sum_{k=0}^{2n-1}{2n-1\choose k}$$
A: $\binom{2n}{2k}$ is the number of subsets of $\{1,\dots,2n\}$ of size $2k$. When you sum these binomial coefficients over all $k$ from $0$ through $n$, you’re counting the number of subsets of $\{1,\dots,2n\}$ whose cardinalities are even. For $n>0$ exactly half of the subsets have even cardinalities, so the sum is $\frac12(2^{2n})=2^{2n-1}$.
Clearly $\{1\}$ has one even subset, $\varnothing$, and one odd subset, $\{1\}$. Suppose that $\{1,\dots,n\}$ has $2^{n-1}$ even and $2^{n-1}$ odd subsets. Now look at the $2^{n+1}$ subsets of $\{1,\dots,n+1\}$. Half of them are $2^n$ subsets of $\{1,\dots,n\}$, of which $2^{n-1}$ are even and $2^{n-1}$ are odd. The other $2^n$ subsets all contain $n+1$. The even ones are obtained by adding $n+1$ to an odd subset of $\{1,\dots,n\}$, so there are $2^{n-1}$ of them. The odd ones are obtained by adding $n+1$ to an even subset of $\{1,\dots,n\}$, so there are $2^{n-1}$ of them as well. Thus, $\{1,\dots,n+1\}$ has $2^{n-1}+2^{n-1}=2^n$ even subsets and the same number of odd subsets.
This does fail for $n=0$, since the empty set has only one subset, itself, and therefore has one even and no odd subsets. In that case $$\sum_{k=0}^n\binom{2n}{2k}=\binom00=1\;.$$
A: from binomial theorem we have
$$\sum_{i=0}^{2m}\binom{2m}{i}x^{i}=(1+x)^{2m}$$
for $x=1$ and $x=-1$ we get
$$\sum_{i=0}^{2m}\binom{2m}{i}=\sum_{k=0}^{2m}\binom{2m}{2k}+\sum_{k=1}^{2m}\binom{2m}{2k-1}=2^{2m}$$
$$\sum_{i=0}^{2m}\binom{2m}{i}(-1)^{i}=\sum_{k=0}^{2m}\binom{2m}{2k}-\sum_{k=1}^{2m}\binom{2m}{2k-1}=0$$
suming these equations we get
$$2\sum_{k=0}^{2m}\binom{2m}{2k}=2^{2m}$$ finally
$$\sum_{k=0}^{2m}\binom{2m}{2k}=2^{2m-1}$$ 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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With $\ds{n \in \mathbb{N}_{\ \geq\ 0}}$:

\begin{align}
\sum_{k = 0}^{n}{2n \choose 2k} & =
\sum_{k = 0}^{2n}{2n \choose k}{1 + \pars{-1}^{k} \over 2}
\\[5mm] & =
{1 \over 2}\sum_{k = 0}^{2n}{2n \choose k}1^{k} +
{1 \over 2}\sum_{k = 0}^{2n}{2n \choose k}\pars{-1}^{k}
\\[5mm] & =
{1 \over 2}\pars{1 + 1}^{2n} + {1 \over 2}\bracks{1 + \pars{-1}}^{2n}
\\[5mm] & =
\bbx{2^{2n - 1} + {1 \over 2}\,\delta_{n0}}
\\ & \
\end{align}
A: Using line integrals: taking $r>1$,
$$
\eqalign{2\pi i\sum_{k=0}^n\binom{2n}{2k}
&= \sum_{k=0}^n\int_{|z|=r}\frac{(z + 1)^{2n}}{z^{2k+1}}\,dz
 = \sum_{k=0}^\infty\int_{|z|=r}\frac{(z + 1)^{2n}}{z^{2k+1}}\,dz
 = \int_{|z|=r}\frac{(z + 1)^{2n}}z\sum_{k=0}^{\infty}\frac1{z^{2k}}\,dz\cr
&= \int_{|z|=r}\frac{(z + 1)^{2n}}z\,\frac1{1 - 1/z^2}\,dz
 = \int_{|z|=r}\frac{z(z + 1)^{2n-1}}{z-1}\,dz = 2\pi i\,2^{2n-1}.
}
$$
