# 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!!

• Perhaps a naive question,but what about induction? – MathematicianByMistake May 30 '15 at 13:58
• Where did the coefficient $2^n$ go right before the line "which is the same as..."? – muaddib May 30 '15 at 13:59
• @muaddib $4^n/2^n=4^m$. – 355durch113 May 30 '15 at 17:31

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 $$\frac{(1+1)^{2m+1}}2 = \frac{2^{2m+1}}2 = 2^{2m} = 4^m$$