The $n$-th term of $16 + 8 + 4 + 2 + \cdots$ is —

With $a=16=2^{4}$ and $r=\tfrac12$, the $n$-th term is $2^{4}\cdot\left(2^{-1}\right)^{n-1}=2^{4-(n-1)}=2^{5-n}$, matching $2^{4}=16$ at $n=1$.