Solving $\sum_{i=0} ^{\log n} i2^i$ 
Solve (or simplify): $$\sum_{i=0} ^{\log n} i2^i$$ (without integrals)

Trying to change the parameter: $j=i2^i$, so since $ 0\le i \le \log n$, then the maximum value for $j$ is when $j=n\log n$. So we get $ \displaystyle \sum_{i=0} ^{\log n} i2^i = \sum_{j=0} ^{n\log n} j$ 
Now I can use the formula to get $ \displaystyle \sum_{j=0} ^{nlog n} j = \Theta ((n \log n)^2)$ but it doesn't seem right when I compare it with a series calculator. 
Maybe did I get the upper bound wrong? 
PS: Here: $\log n = \log _2 n$
 A: Note that
\begin{align}
\sum_{i=1}^{n}i \cdot 2^{i} &= 2+(4+4)+(8+8+8)+\ldots+(\underbrace{ 2^{n}+2^{n}+\ldots+2^{n}}_{n\text{ times}})\\
&=(2+4+8+\ldots+2^{n-1}+2^{n})+(4+8+\ldots+2^{n-1}+2^{n})+(8+\ldots\\
&\quad+2^{n-1}+2^{n})+\ldots+(2^{n-2}+2^{n-1}+2^{n})+(2^{n-1}+2^{n})+2^{n}\\
&=2(1+2+4+\ldots+2^{n-2}+2^{n-1})+4(1+2+4+\ldots+2^{n-3}+2^{n-2})+8(1+\ldots\\
& \quad+2^{n-3})+\ldots+2^{n-2}(1+2+4)+2^{n-1}(1+2)+2^{n}\\
&=2(2^{n}-1)+4(2^{n-1}-1)+8(2^{n-2}-1)+\ldots+2^{n-2}(2^{3}-1)+2^{n-1}(2^{2}-1)+2^{n}\\
&= (n-1)2^{n+1}+2^{n}-2-4-\ldots-2^{n-1}=(n-1)2^{n+1}+2^{n}-2(2^{n-1}-1)\\
&= (n-1)2^{n+1}+2\\
&= 2(2^n(n-1) + 1)
\end{align}
thus, assuming that $\log n = \log_2 n$, you get
$$\sum_{i=0} ^{\log n} i2^i = 2(1+2^{\log n}(\log n-1)) = 2(1+n(\log n-1))$$
A: The issue is that the change in parameter is done poorly. If you write out the first few terms, you see that your parameter change changes the sum.
Try the following instead: Let $f_k(x)=\sum_{i=0}^k x^{i+1}$. Then $$f'_k(x)=\sum_{i=0}^k (i+1)x^i=\sum_{i=0}^k ix^i+\sum_{i=0}^k x^i$$
Substitute in the explicit formula for $\sum_{i=0}^k x^i$, calculate the derivative on the LHS, and simplify. Then plug in $\log(n)$ and $2$
A: Take into account that $$\sum_{i=0} ^{x} i2^i = x 2^{x + 1} - 2^{x + 1} + 2$$ and put $x = \log_{2} n$ to get $$2 (\log_{2} n - 1) n + 2.$$
