How is this sequence in arithmetico-geometric progression (AGP)?

Question

I've started learning about APs, GPs and AGPs, and, while the sequence $\{x,2x^2,3x^3,...\}$ seems like a textbook example of an arithmetico-geometric progression -- since it's the product of the AP $\{1,2,3,...\}$ and the GP $\{x,x^2,x^3,...\}$, it's clearly not of the form $\{a, (a+d)r, (a+2d)r^2,...\}$.

Motivation: I'm trying to find the sum to $n$ terms of this sequence, but I'm not sure exactly what $a, d$ and $r$ are supposed to be (where $a$ is the initial term, $d$ is the "common difference" and $r$ is the common ratio). I know that I can factor out the $x$ and then I just need to calculate the sum of $\{1,2x,3x^2, ..., nx^{n-1}\}$ ( where $a=1, d=1, r=x$) but I want to actually know what they are for the original sequence! Any help would be very appreciated.

Answer 1

If I properly understand, you want to compute $$S_n=\sum_{i=1}^n i x^i$$ Just rewrite $$S_n=\sum_{i=1}^n i x^i=x\sum_{i=1}^n i x^{i-1}=x \left(\sum_{i=1}^n x^{i} \right)'$$ where you find the derivative of a geometric progression.

So, $$\sum_{i=1}^n x^{i} =\frac{x \left(x^n-1\right)}{x-1}$$ $$\left(\sum_{i=1}^n x^{i} \right)'=\frac{(n (x-1)-1) x^n+1}{(x-1)^2}$$ $$S_n=x\frac{(n (x-1)-1) x^n+1}{(x-1)^2}$$

Answer 2

to solve write it like this :
$S= x,2x^2,3x^3,\dots, T_{n-1} , T_n$
multiply it by the common ratio of gp i.e. $x$ to get:
$xS= \quad x^2,2x^3,3x^4,\dots$
Notice how I wrote it by shifting it by one term to the right

Do the difference then solve the resulting series.