Q.Find the sum to $n$ terms of the arithmetic-geometric series $2 + 5\left(\dfrac13\right) + 8\left(\dfrac13\right)^2 + 11\left(\dfrac13\right)^3 + \ldots$

[!TLDR] A.G.P. with $a=2,d=3,r=\tfrac13$; substitute into the boxed formula and simplify. [!ANSWER] $S_n=\d …

Here $a=2,d=3,r=\tfrac13$. Substituting into $S_n=\dfrac{a}{1-r}+\dfrac{dr(1-r^{n-1})}{(1-r)^2}-\dfrac{[a+(n-1)d]r^n}{1-r}$: $\dfrac{a}{1-r}=\dfrac{2}{2/3}=3$; with $dr=1$ and $(1-r)^2=\dfrac49$, the middle term is $\dfrac{1-(1/3)^{n-1}}{4/9}=\dfrac94\left[1-\left(\tfrac13\right)^{n-1}\right]$; the last term, with $a+(n-1)d=3n-1$, is $\dfrac{(3n-1)(1/3)^n}{2/3}=\dfrac{(3n-1)}{2}\left(\tfrac13\right)^{n-1}$. Combining and collecting the $\left(\tfrac13\right)^{n-1}$ terms: $S_n=3+\dfrac94-\left[\dfrac94+\dfrac{3n-1}{2}\right]\left(\tfrac13\right)^{n-1}=\dfrac{21}{4}-\dfrac{6n+7}{4}\left(\tfrac13\right)^{n-1}=\dfrac{21-(6n+7)(1/3)^{n …

Substitute $a,d,r$ into the boxed A.G.P. formula, then collect the two $\left(\frac13\right)^{n-1}$-bearing terms together and s …

Leaving the answer as two separate un-combined fractional terms (rather than simplifying to one closed form) makes later verification mu …