What is the result of this sum, for big values of n?

Question

The sum is the following: $$\sum_{i=0}^{n-1} \left[i\frac{2^i}{2^n-1}\right]$$

I know that the result of this sum, for big values of $n$, should be $n-2$, but I am not aware of the procedure used to get to this solution.

Answer 1

Hint

You have

$$S_n(x) = \sum_{i=0}^{n-1} x^i = \frac{1-x^n}{1-x}$$

Hence

$$\sum_{i=0}^{n-1} i x^i = x S_n^\prime(x)$$

and finally

$$\sum_{i=0}^{n-1} i*\frac{2^i}{2^n-1} = \frac{2}{2^n-1}S_n^\prime(2)$$