Question on equations and algorithms This question is regarding discreet math. I need to check if this equation is right: 
$\sum_{i = 1}^{n} 7^i log_2 (i) = \Theta(7^n log_2 (n))$ 
How can I prove that? I am new to this so I am sorry if I did something wrong for this question. I am having a lot of difficulties with this one in particular and I have no idea where to start. I have to resort to this because I cannot get an answer from my professor. 
 A: First of all, $\log_2 i$ is always smaller than $\log_2 n$ in your sum. So you have:
$$
\sum_{i=1}^n 7^i \log_2 i \leq \log_2 n \sum_{i=1}^n 7^i
$$
Then, using the sum of terms of a geometry series, you get
$$
\sum_{i=1}^n 7^i \log_2 i \leq \log_2 n \left(\frac{7^{n+1} - 7}{6}\right).
$$
This is enough to show that your sum is a $\mathcal{O}\left(\log_2 n 7^n\right)$. Now you just have to find a matching lower-bound. Hope that helps...
A: It's sufficient to investigate
$$S_n:=\sum_{k=1}^n 7^k\log k.$$
Now divide it into two parts :
$$S_n=\sum_{k=1}^{n-\lfloor\log n\rfloor}+\sum_{k=n-\lfloor\log n\rfloor+1}^n=S_n^1+S_n^2.$$
It's trivial that
$$S_n^1\leqslant 7^{n+2-\log n}\log n=\mathcal{O}\left(\frac{7^n\log n}{n^{\log 7}}\right).$$
On the other hand,
\begin{align*}
S_n^2& =\sum_{k=1}^{\lfloor\log n\rfloor}7^{n+1-k}\log(n+1-k)\\
& =7^{n+1}\sum_{k=1}^{\lfloor\log n\rfloor}\frac{\log(n+1)+o(1)}{7^k}\\
& =\frac{1}{6}\cdot 7^{n+1}\log n(1+o(1)).
\end{align*}
Thus $S_n=\frac{1}{6}\cdot 7^{n+1}\log n(1+o(1)).$
Then as $n\to \infty$,
$$\sum_{k=1}^n 7^k\log_2 k\sim\frac{7^{n+1}\log n}{6\log 2}.$$
