How can I simplify $\sum_{k=0}^n a \cdot k \cdot q^k$?

Question

So I recently found out about geometrical series. However now I've got this formula:

$\sum_{k=0}^{n-1} \frac 1 3 k \cdot {\frac 2 3}^k$ is there also a way to simplify this?

Answer 1

Let $S$ be the desired sum.

$$S=q+2q^2+3q^3+\dots+nq^n$$

Multiply it by $(1-q)$.

$$(1-q)S=(\color{red}q+\color{blue}{2q^2}+\color{red}{3q^3}+\dots+nq^n)-(\color{blue}{q^2}+\color{red}{2q^3}+\color{blue}{3q^4}+\dots+nq^{n+1})$$

Combine like terms. (notice the minus sign on the second set of values)

$$=\color{red}q+\color{blue}{[2-1]q^2}+\color{red}{[3-2]q^3}+\dots+[n-(n-1)]q^n-nq^{n+1}$$

$$=q(1+q+q^2+\dots+q^n)-nq^{n+1}$$

Let $G$ be the geometric sum.

$$(1-q)S=q(G-nq^n)$$

Solve for $S$.

$$S=\frac{q(G-nq^n)}{1-q}$$

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The geometric sum:

$$G=1+q+q^2+\dots+q^n$$

Multiply it by $(1-q)$.

$$(1-q)G=(\color{red}1+\color{blue}q+\color{red}{q^2}+\dots+q^n)-(\color{blue}q+\color{red}{q^2}+\color{blue}{q^3}+\dots+q^{n+1})$$

Combine like terms. (notice the minus sign on the second set of values)

$$=\color{red}1+\color{blue}{[1-1]q}+\color{red}{[1-1]q^2}+\dots+[1-1]q^n-q^{n+1}$$

$$(1-q)G=1-q^{n+1}$$

Solve for $G$.

$$G=\frac{1-q^{n+1}}{1-q}$$