discrete mathematics - Prove $ sum_{k=0}^n k4^k = frac 49((3n-1)4^n + 1) $ by induction

Prove that for every position integer $n$ that

$$ \sum_{k=0}^n k4^k = \frac 49((3n-1)4^n + 1) $$

Proof: Let $P(n)$ denote the above statement.

Base case: $n=1$ : Note that $$ \sum_{k=1}^1 k4^k = \frac 49((3(1)-1)4^{(1)} + 1) $$

$\frac 49((3(1)-1)4^{(1)} + 1) = \frac49((2)4+1) = \frac49(8+1) = \frac 49(9) = 4$

$k4^k = (1)4^{(1)} = 4$

So P(1) holds.

Inductive Step: Let $s\ge1$. Assume P(s), so

$$ \sum_{k=1}^s k4^k = \frac 49((3s-1)4^s + 1) $$

Note

$$ \sum_{k=1}^{s+1} k4^k = \sum_{k=1}^{s} k4^k + (s+1)4^{s+1} $$

and by inductive hypothesis:

**

$$ \frac 49((3s-1)4^s + 1) + (s+1)4^{s+1} $$ **

I'm afraid I'm stuck after this point. I know my endpoint needs to be:

$$ \sum_{k=1}^{s+1} k4^k = \frac 49((3(s+1)-1)4^{s+1} + 1) $$

but I don't know how to get from the asterisks to the above. Any help would be greatly appreciated.