Q.Find the sum of the first $5$ terms of the arithmetic-geometric series $1 + 3\left(\dfrac12\right) + 5\left(\dfrac12\right)^2 + 7\left(\dfrac12\right)^3 + 9\left(\dfrac12\right)^4$.

[!TLDR] Add the 5 terms directly (an A.G.P. with $a=1,d=2,r=\tfrac12$), and cross-check with the boxed A.G.P. form …

The $n$th term is $(2n-1)\left(\tfrac12\right)^{n-1}$, an A.G.P. with A.P. part $1,3,5,7,9$ ($a=1,d=2$) and G.P. part ratio $r=\tfrac12$. Adding the 5 terms directly: $1 + 1.5 + 1.25 + 0.875 + 0.5625 = 5.1875=\dfrac{83}{16}$. As a cross-check using the boxed A.G.P. formula $S_n=\dfrac{a}{1-r}+\dfrac{dr(1-r^{n-1})}{(1-r)^2}-\dfrac{[a+(n-1)d]r^n}{1-r}$ with $a=1,d=2,r=\tfrac12,n=5$: $\dfrac{a}{1-r}=\dfrac{1}{1/2}=2$; $\dfrac{dr(1-r^{4})}{(1-r)^2}=\dfrac{1\cdot(1-\frac{1}{16})}{1/4}=\dfrac{15/16}{1/4}=\dfrac{15}{4}$; $\dfrac{[a+4d …

Either add the 5 numeric terms directly, or substitute $a,d,r,n$ into the boxed A.G.P. sum f …

A common slip when using the formula is using $r^{n}$ where $r^{n-1}$ is needed (or vice versa) in the middle term, since the two exponents …