Sigma notation breakdown What does the sum below evaluate to?
$$\sum_{i=1}^n 4^i*i $$
I know the rule for a sum of a geometric series:
$$\sum_{i=1}^n a\cdot r^{i-1}=\frac{a(r^n-1)}{r-1} $$
But I cant seeem to apply to this formula because its to the power of i
and there is also the *i factor
 A: Such a series in which the terms are product of GP and an AP is called an AGP. 
The method for evaluating such sums is simple. To determine a formula for this sum without using derivatives (as suggested in the comment) you can use the following method. 
Let $$\sum_{i=1}^n 4^i*i = S$$
Or S=$$\sum_{i=0}^{n-1}4^{i+1}*{(i+1)} $$
Then $$4S=\sum_{i=1}^n4^{i+1}*i $$
Implies $$3S= 1+ n4^{n+1} - \sum_{i=0}^{n}4^i $$
And done. 

EDIT:
To clarify how I got 3S, I will list S and 4S.
S= $1*4+2*4^2+3*4^3+...+n4^n$
4S=        $1*4^2+2*4^3+...+(n-1)4^n + n4^{n+1}$
On subtracting we get-
$3S= -4-4^2-4^3-...-4^n + n4^{n+1}$
Or $$3S= 1+ n4^{n+1} - \sum_{i=0}^{n}4^i $$
A: $$\sum_{i=1}^n 4^i*i $$ means that sum all the terms from $i=1$ to $i=n$.
So:
$\sum_{i=1}^n 4^i*i =4^1* (1)+4^2* (2)+4^3* (3)+...…+4^n* (n)$ so add all terms one by one until you reach $n$ while replacing $i$ by the term number.
A: The formula $3S= 1+ n4^{n+1} - \sum_{i=0}^{n}4^i$ (where $S=\sum_{i=1}^n 4^i\cdot i$) from @Random-15's answer simplifies to
$$S=\frac{(3n-1)4^{n+1}+4}9$$ which does produce correct answers $4, 36$ and $228$ for $n=1,2,3,$ respectively. 
Now, knowing the formula let me try to prove it by induction. 
When $n=1$ we have $\sum_{i=1}^1 4^i\cdot i = 4^1\cdot1=4$ and 
$\frac{(3\cdot1-1)4^{1+1}+4}9=\frac{(2)4^2+4}9=\frac{36}9=4$, so the formula 
$S=\frac{(3n-1)4^{n+1}+4}9$ is correct when $n=1$. 
Assume formula is correct for some $n\ge1$. Prove it for $n+1$. We have 
$$\sum_{i=1}^{n+1} 4^i\cdot i=\sum_{i=1}^n 4^i\cdot i +4^{n+1}\cdot(n+1)=$$ (by induction hypothesis)
$$=\frac{(3n-1)4^{n+1}+4}9+4^{n+1}\cdot(n+1)= 
\frac{(3n-1+9n+9)4^{n+1}+4}9=$$
$$=\frac{(12n+12-4)4^{n+1}+4}9=\frac{4(3n+3-1)4^{n+1}+4}9=$$
$$=\frac{\bigl(3(n+1)-1\bigr)4^{n+2}+4}9$$ which completes the proof. 
Note. Wolfram alpha also gives a similar formula:
$\sum_{i=1}^n 4^i\cdot i = \frac49 (3\cdot 4^n\cdot n - 4^n + 1)$. 
