How can I compute sum of $i 4^i$?

Question

How can I compute $$\sum_{i=0}^n i 4^i$$ this equation? What is the way?

Answer 1

\begin{align} \sum_{i=0}^n i 4^i &= 4\sum_{i=0}^n i 4^{i-1}\\ &=4 \sum_{i=0}^n \frac{d}{dx}x^i|_{x=4} \\ &=4 \frac{d}{dx}\left[\sum_{i=0}^n x^i \right]_{x=4} \\ &= 4 \left[\frac{d}{dx}\left[\frac{x^{n+1}-1}{x-1}\right]\right]_{x=4}\\ \end{align}

Answer 2

Hint

The given sequence is an arithmetico Geometric series. Let $S$ be sum of arithmetico geometric progression then $$S=\frac {a-[a+(l-1)d]r^l}{1-r}+\frac {dr(1-r^{l-1})}{(1-r)^2}$$

For your question $a=0,r=4,d=1,l=n+1$