Compute $\binom n0 + 2\binom n1 + \binom n2 + 2\binom n3+...$

Question

Compute $\binom n0 + 2\binom n1 + \binom n2 + 2\binom n3+...$

Well, I think $a_1,a_3,...$ are $2^{n-1}$. and $a_2,a_4,..$ are $2^n$. Thus, the answer is $2^{n-1} + 2^n$.

What's your opinion guys?

Answer 1

Your conclusion is right for $n>0$, but your reasoning is wrong (or has a typo): both the even- and the odd-numbered terms sum to $2^{n-1}$ (assuming that $n>0$), so the total is

$$3\cdot 2^{n-1}=2\cdot2^{n-1}+2^{n-1}=2^n+2^{n-1}\;.$$

That is, you have each odd-numbered term twice and each even-numbered term only once. Alternatively, the sum is

$$\sum_{k\ge 0}\binom{n}k+\sum_{k\ge 0}\binom{n}{2k+1}=2^n+2^{n-1}=3\cdot2^{n-1}\;.$$

Note that $n=0$ is a special case that has to be handled separately: in that case the sum is simply $1$.