According to book the answer is: $2n+1$.
I have no idea where it came from. I have tried to use the n-th term of a geometric sequence formula, but without any luck. Maybe I am just tired - any help would be greatly appreciated.
$$ 2^{2n}=2^{n-1} $$
By the way, given sequence is just a numerator that is part of a limit exercise.
Update: The above $n$, as I said, is part of the limit exercise, where $n\rightarrow\infty$. I have confused the $n$ from $\lim$ with then $n$ from the n-th term of a geometric sequence formula.
$$\begin{align} 2^{2n}=2^{k-1}\\ 2n=k-1\\ k = 2n + 1 \end{align}$$
where $k$ represents the number of terms. Duck debugging works as always and the Earth spins again.
There is $2n-0+1$ terms since the sum start at $1= 2^0$ and end at $2^{2n}$
In general if you have $$u_k+\cdots +u_m$$ then the number of term is $\color{blue}{m-k+1}.$