1
$\begingroup$

I'm having a hard time proving:

$$ \sum_{k = 0}^n {{2n + 1}\choose{2k + 1}} = 4^n $$

The initial step is rather simple to prove, but I seem to miss something important in the induction step ($n \Rightarrow n + 1$).

This is what I got so far:

$$ \begin{alignat*}{3} &\sum_{k = 0}^{n + 1} {{2(n + 1) + 1}\choose{2k + 1}} \\ = &\sum_{k = 0}^{n + 1} {{2n + 3}\choose{2k + 1}} \\ = &\sum_{k = 0}^{n} {{2n + 3}\choose{2k + 1}} + {{2n + 3}\choose{2n + 3}} \\ = &\sum_{k = 0}^{n} {{2n + 3}\choose{2k + 1}} + 1 \\ \end{alignat*} $$

I cannot find a way to use the induction hypothesis, since the binomial coefficient inside the sum contains the term $2n + 3$ instead of $2n + 1$.

How can I proceed from here? Any tips?

  • 0
    It's much easier without induction. :) But you might be able to use that $$\binom{2n+3}{2k+1}=\binom{2n+1}{2k-1}+2\binom{2n+1}{2k}+\binom{2n+1}{2k+1}$$ Still not seeing how to remove the $\binom{2n+1}{2k}$ terms without appealing to the same theorem that would let you avoid the induction in the first place. – 2017-01-25
  • 1
    Since $$\binom{2n+1}{2k+1} = \binom{2n}{2k+1}+\binom{2n}{2k}$$ the given sum equals $$\sum_{k=0}^{2n}\binom{2n}{k} = 2^{2n}.$$ – 2017-01-25
  • 0
    @ThomasAndrews How would you prove without induction look like? – 2017-01-25
  • 0
    It's basically what Jack wrote, @Herickson. – 2017-01-25
  • 0
    @ThomasAndrews Okay so: $$ \sum_{k = 0}^{n} {{2n + 1}\choose{2k + 1}} = \sum_{k = 0}^{n} {{2n}\choose{2k + 1}} + {{2n}\choose{2k}}$$ This contains all the even and odd terms from $0$ up to $2n + 1$, but since ${{2n}\choose{2n + 1}} = 0$ we are able to rewrite the sum into $\sum_{k = 0}^{2n} {{2n}\choose{k}}$, right? – 2017-01-25
  • 0
    Right, and that is $2^{2n}=4^n$. @Herickson – 2017-01-25
  • 0
    Another proof without induction: The sum in question is the total number of odd-sized subsets of $\{1,2,\dots,2n,2n+1\}$. Such a subset can be formed by arbitrarily choosing a subset of $\{1,2,\dots,2n\}$ and then adjoining $2n+1$ iff the chosen subset had even cardinality. So the number of possibilities is the number of subsets of $\{1,2,\dots,2n\}$, namely $2^{2n}$. – 2017-01-25
  • 0
    Great way to think about it. Outstanding intuition! – 2017-01-25

1 Answers 1

1

Here's an inductive proof.

Using $\binom{a+1}{b}=\binom{a}{b}+\binom{a}{b-1}$ twice, we get:

$$\begin{align}\binom{2n+3}{2k+1}&=\binom{2n+2}{2k+1} + \binom{2n+2}{2k}\\ &=\binom{2n+1}{2k+1}+2\binom{2n+1}{2k}+\binom{2n+1}{2k-1}\\ &=\binom{2n+1}{2k+1}+2\binom{2n+1}{2(n-k)+1}+\binom{2n+1}{2k-1} \end{align}$$

The last equality because $\binom{a}{b}=\binom{a}{a-b}$.

Now, how many times does $\binom{2n+1}{2j+1}$ occur in:

$$\sum_{k=0}^{n+1}\binom{2n+3}{2k+1}=\sum_{k=0}^{n+1}\left(\binom{2n+1}{2k+1}+2\binom{2n+1}{2(n-k)+1}+\binom{2n+1}{2k-1}\right)?$$

It occurs once when $k=j$ once when $k=j+1$ and twice when $n-k=j$.

  • 0
    Thank you! I'll need some time to fully think through this, but will happily accept afterwards :) – 2017-01-25
  • 0
    It's clear to see that ${{2n + 1}\choose{2j + 1}}$ occurs exactly four times in the last sum. But how can we be sure, that the sum does not contain *other* terms and is therefore greater than $4 \cdot \, \sum_{k = 0}^{n} {{2n + 1}\choose{2j + 1}}$? – 2017-01-25
  • 0
    Well, every term in the rewritten sum is of the form $\binom{2n+1}{2j+1}$. But some of the values of $j$ are outside the range $0,\dots,n$. And those terms don't occur four times. But why would that not matter? – 2017-01-25
  • 0
    You are right, I was just curious. Thanks a lot! – 2017-01-25
  • 0
    @ThomasAndrews: Elegant argumentation. (+1) But, shouldn't we also consider the boundary values explicitly? – 2017-01-25