3
$\begingroup$

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!!

$\endgroup$
  • $\begingroup$ Perhaps a naive question,but what about induction? $\endgroup$ – MathematicianByMistake May 30 '15 at 13:58
  • $\begingroup$ Where did the coefficient $2^n$ go right before the line "which is the same as..."? $\endgroup$ – muaddib May 30 '15 at 13:59
  • $\begingroup$ @muaddib $4^n/2^n=4^m$. $\endgroup$ – 355durch113 May 30 '15 at 17:31
10
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I get it now. Thanks :). $\endgroup$ – Jazz May 30 '15 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.