Binomal theorem show that $\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\dots=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\dots=2^{n-1}$

Question

I'm having some trouble with this question

Show that $$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\dots=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\dots=2^{n-1}$$

Attempt:

Expanding $(1+1)^n=2^n$ $$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\binom{n}{4}+\dots+\binom{n}{n}=2^n$$

Expanding $(1-1)^n=0$ $$\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\binom{n}{4}+\dots+(-1)^n\binom{n}{n}=0$$

And I'm stuck here. The solution says "Adding these two equalities yields the required one".

I can see how the terms will cancel out,however I don't see how $2^n+0 = 2^{n-1}$

Answer 1

Add both equalities you got:

$$2\left(\binom n0+\binom n2+\ldots\right)=2^n+0$$

and now divide by two. Finally, oberve the symmetry between binomial coefficients $\;\binom nk\;$ with even $\;k\;$ and with odd $\;k\;$

Answer 2

Also you can subtract equation 2) from 1) to get second part.