How to find the closed form of the summation below without changing the lower and upper bound of summation?

Question

The lower bound of summation is i=0 and the upper bound of summation is log(n) - 1 (log is base 2).

Answer 1

$f(x)=\sum_{k=0}^mx^k=\frac{x^{m+1}-1}{x-1}$, so $f'(x)=\sum_{k=0}^mkx^{k-1}=\frac{(m+1)x^m(x-1)-x^{m+1}+1}{(x-1)^2}$ For your question you want $A=2f'(2)$ where $m=log_2(n)-1$. So $A=log_2(n)\times n-2n+2$.

This derivation is for $log_2(n)$ an integer or $n=2^j$ for some integer $j$. Otherwise you need to use a greatest integer function for the sum limit.