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

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:
$\sum_{i=1}^n 4^i\cdot i = \frac49 (3\cdot 4^n\cdot n - 4^n + 1)$. 
