Sum of binomial coefficients from $\binom{2m+1}{0}$ to $ \binom{2m+1}{m} $ Prove that: $\displaystyle 4^m = \binom{2m+1}{0}+\binom{2m+1}{1}+\binom{2m+1}{2}+\ldots + \binom{2m+1}{m} $

From the Binomial Theorem:
$\displaystyle (a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$
If $\displaystyle a = b = 2$, then we have
$\displaystyle 4^n = \sum_{k=0}^{n}\binom{n}{k}2^{n-k}2^k = 2^n\sum_{k=0}^{n}\binom{n}{k} = 2^n\cdot 2^n = 4^n$
Which is the same as $\displaystyle \sum_{k=0}^{2m}\binom{2m}{k}$ if $a = b = 1$ and $n =2m$ (with $4^m$ though).
What I'm having trouble with is the expansion. I've tried two things:
1) $\displaystyle \binom{2m}{0} = \binom{2m+1}{0}$ and applying the Pascal Identity to the remaining binomial coefficients:
\begin{align*}\sum_{k=0}^{2m}\binom{2m}{k} &= \binom{2m}{0}+\binom{2m}{1}+\binom{2m}{2}+\binom{2m}{3}+ \ldots \\
&= \binom{2m+1}{0} + \binom{2m+1}{2} + \binom{2m+1}{4} + \ldots
\end{align*}
2) Applying the Pascal Identity from the 1st binomial coefficient:
\begin{align*}\sum_{k=0}^{2m}\binom{2m}{k} &= \binom{2m}{0}+\binom{2m}{1}+\binom{2m}{2}+\binom{2m}{3}+ \ldots \\
&= \binom{2m+1}{1} + \binom{2m+1}{3} + \binom{2m+1}{5} + \ldots\end{align*}
I don't know how to get the coefficients I'm asked to have.
Thanks!!

Your sum is the first half of the binomial expansion of $(1+1)^{2m+1}$ -- and since $\binom{2m+1}n = \binom{2m+1}{2m+1-n}$, the sum is equal to the other half, so the sum is
$$\frac{(1+1)^{2m+1}}2 = \frac{2^{2m+1}}2 = 2^{2m} = 4^m $$
