7.0 Arithmetico-Geometric Series (A.G.S.)

A series whose each term is formed by multiplying corresponding terms of an A.P. and a G.P., is called an Arithmetic-Geometric series.

For example: $1 + 3x + 5{x^2} + 7{x^3} + ...$

Here, $1,3,5,7...$ are in A.P. and $1,x,{x^2},{x^3},...$ are in G.P.

Proof: $${S_n} = a + (a + d)r + (a + 2d){r^2} + ... + \left[ {a + (n - 1)d} \right]{r^{n - 1}}...(1)$$

Multiplying both sides of $(1)$ by common ratio $r$ and write as follows

$$r{S_n} = 0 + ar + (a + d){r^2} + (a + 2d){r^3} + ... + \left[ {a + (n - 1)d} \right]{r^n}...(2)$$

Subtracting $(2)$ from $(1)$, we get

$$\begin{equation} \begin{aligned} {S_n}(1 - r) = a + [dr + d{r^2} + ... + d{r^{n - 1}}] - [a + (n - 1)d]{r^n} \\ {\text{ = }}a + d[r + {r^2} + {r^3} + ...{r^{n - 1}}] - [a + (n - 1)d]{r^n} \\ {\text{ = }}a + \frac{{dr(1 - {r^{n - 1}})}}{{1 - r}} - [a + (n - 1)d]{r^n} \\ \Rightarrow {S_n} = \frac{a}{{1 - r}} + \frac{{dr(1 - {r^{n - 1}})}}{{1 - r}} - \frac{{\left[ {a + (n - 1)d} \right]{r^n}}}{{1 - r}} \\\end{aligned} \end{equation} $$

Solution: $${S_\infty } = 1 + 2x + 3{x^2} + 4{x^3} + ...\infty ...(1)$$ Multiply both sides by $x$ and write the equation as follows $$x{S_\infty } = 0 + x + 2{x^2} + 3{x^3} + 4{x^4} + ...\infty \quad ...(1)$$ Subtracting $(2)$ from $(1)$, we get $$\begin{equation} \begin{aligned} (1 - x){S_\infty } = 1 + x + {x^2} + {x^3} + ...\infty \\ {\text{ = }}\frac{1}{{1 - x}} \\ \Rightarrow {S_\infty } = \frac{1}{{{{\left( {1 - x} \right)}^2}}} \\\end{aligned} \end{equation} $$