Asymptotic – Did I try \$ sum_ {i = 1} ^ { lg n} 2 ^ {i-1} = Theta (n lg n) \$?

I have an exercise problem and I do not know why your answer is like that.
$$sum_ {i = 1} ^ { lg n} 2 ^ {i-1} in Theta (2 ^ { lg n}) = Theta (n).$$

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 test,
$$sum_ {i = 1} ^ {g (n)} f (i) le sum_ {i = 1} ^ {g (n)} f (g (n)) = g (n) cdot f (g (n)).$$

My thinking with this question, $$f (n) = 2 ^ {i-1}$$ Y $$g (n) = lg n$$. Then the result would be $$lg n cdot 2 ^ { lg n – 1}$$.

Can someone indicate where I made an error? (I'm not asking about my rating, but tell me where I think wrong)