Q.Find the sum to $n$ terms of the arithmetic-geometric series $1 + 4x + 7x^2 + 10x^3 + \ldots$ (for $x \ne 1$).

[!TLDR] This is an A.G.P. with $a=1,d=3$, ratio $x$; substitute into the boxed A.G.P. formula. [!ANSWER] $S_n=\dfrac{1}{1-x}+\dfrac{3x(1-x^{n-1})}{(1-x)^2}-\dfrac{(3n-2)x^n}{1-x}$.

The series has A.P. part $1,4,7,10,\ldots$ ($a=1,d=3$) and G.P. ratio $x$, so its $n$th term is $[1+(n-1)3]x^{n-1}=(3n-2)x^{n-1}$; in particular $a+(n-1)d=1+3(n-1)=3n-2$. Substituting $a=1,d=3,r=x$ into the boxed formula of Section 7, $S_n=\dfrac{a}{1-r}+\dfrac{dr(1-r^{n-1})}{(1-r)^2}-\dfrac{[a+(n-1)d]r^n}{1-r}$, gives $S_n=\dfrac{1}{1-x}+\dfrac{3x(1-x^{n-1})}{(1-x)^2}-\dfrac{(3n-2)x^n}{1-x}$. As a check at $n=1$: the formula should give just the first term $1$; substituting $n=1$ makes the middle term's $(1-x^0)=(1-1)=0$ and the last term becomes $\dfrac{1\cdot x}{1-x}$, and $\dfrac{1}{1-x}-\dfrac{x}{1-x}=\dfrac{1-x}{1-x}=1$ ✓. [!ANSWER] $S_n=\dfrac{1}{1-x}+\dfrac{3x(1-x^{n-1})}{(1-x)^2}-\dfrac{(3n-2)x^n}{1-x}$.

Identify $a,d$ from the A.P. part and $r$ from the ratio, then substitute directly into the boxed A.G.P. sum formula of Section 7; sanity-check at $n=1$.

Confusing which exponent is $n-1$ and which is $n$ in the two different $x$-power terms of the formula is the most common error.