Prove the equality

Question

Prove the equality $\sum\limits_{k=0}^{n-1}{4n\choose 4k+1}=2^{4n-2}$

$$\sum\limits_{k=0}^{n-1}{4n\choose 4k+1}=2^{4n-2}=\frac{1}{2}\sum\limits_{k=0}^{n-1}{4n\choose 4k+1}+\frac{1}{2}\sum\limits_{k=0}^{n-1}{4n\choose 4n-4k-3}$$

I am stuck here.

Could someone give a hint on this?

Answer 1

Since $${4n \choose 4k+1}={4n\choose 4(n-k)-1}$$ we have $$\sum_{k=0}^{n-1}{4n \choose 4k+1}=\sum_{k=0}^{n-1}{4n\choose 4(n-k)-1}=\sum_{k=0}^{n-1}{4n\choose 4k+3}$$

We have $(1-1)^{4n}=0$, which implies, from the binomial theorem $$\sum_{k=0}^{2n}{4n\choose 2k}=\sum_{k=1}^{2n}{4n\choose 2k-1}=\sum_{k=0}^{n-1}{4n \choose 4k+1}+\sum_{k=0}^{n-1}{4n\choose 4k+3}=2\sum_{k=0}^{n-1}{4n \choose 4k+1}\tag{1}$$ From the binomial theorem, again we get $$2^{4n}=(1+1)^{4n}=\sum_{k=0}^{2n}{4n\choose 2k}+\sum_{k=1}^{2n}{4n\choose 2k-1}=2\sum_{k=0}^{2n}{4n\choose 2k}\tag{2}$$ Thus, from ($1$) and ($2$) $$\sum_{k=0}^{n-1}{4n \choose 4k+1}=\frac{2^{4n}}{4}=2^{4n-2}$$

Answer 2

As a hint, try this: $$\begin{align} \sum_{k=0}^{n-1}\binom{4n}{4k+1}&=\sum_{k=0}^{n-1}\binom{4k}{2k+1}\\ &=\sum_{k=0}^{n-1}\left[\binom{4n-1}{2k}+\binom{4n-1}{2k+1}\right]\\ &=\frac{1}{2}\sum_{k=0}^{n-1}\binom{4n-1}{k}\\ &= \frac{1}{2}\,2^{4n-1} = 2^{4n-2} \end{align}$$ Of course, you'll have to convince yourself that each step is valid, but that's not hard to do after looking at an example, say for $n=4$.