## Have I proved $\sum_{i=1}^{\lg n} 2^{i-1} = \Theta(n\lg n)$?

I have an exercise problem and don’t know why its answer is like this. $$\sum_{i=1}^{\lg n} 2^{i-1} \in \Theta(2^{\lg n}) = \Theta(n).$$

Regarding this equation, I think it would be, $$\sum_{i=1}^{\lg n} 2^{i-1} \in \Theta(\lg n \cdot 2^{\lg n}) = \Theta(\lg n \cdot n),$$ with the proof, $$\sum_{i=1}^{g(n)}f(i) \le \sum_{i=1}^{g(n)}f(g(n)) = g(n)\cdot f(g(n)).$$

Can anyone point out where I made a mistake? (I am not asking grading mine, but tell me where I mis-think)