Solve following summation using Binomial theorem

Question

I apologise in advance for posting another one of these homework assignment-questions; this one is pissing me off.

The question is: solve $\sum_{k=0}^n {{n \choose k}} k 4^k $ using the binomial theorem.

The deadline for the homework assignment is already long past (it wasn't mandatory); I just really want to figure out what the deal is here. Any tips are very welcome!

Answer 1

$$ \begin{align} \sum_{k=0}^n\binom{n}{k}k4^k &=4n\sum_{k=0}^n\binom{n-1}{k-1}4^{k-1}\\ &=4n5^{n-1} \end{align} $$ In general $$ \binom{n}{k}k(k-1)\cdots(k-j+1)=\binom{n-j}{k-j}n(n-1)\cdots(n-j+1) $$

Answer 2

Making the problem more general, consider $$\sum_{k=0}^n {{n \choose k}} k\, x^k=x\sum_{k=0}^n {{n \choose k}} k\, x^{k-1}=x\left(\sum_{k=0}^n {{n \choose k}} \, x^{k}\right)'$$ and $$\sum_{k=0}^n {{n \choose k}} \, x^{k}=(1+x)^n$$ $$\left(\sum_{k=0}^n {{n \choose k}} \, x^{k}\right)'=n\,(1+x)^{n-1}$$ which finally makes$$\sum_{k=0}^n {{n \choose k}} k\, x^k=n\,x\,(1+x)^{n-1}$$ Now, make $x=4$.