progress in progressions

Recall that the sum of the first n terms of a geometric progression is $\dfrac{a\left(r^n - 1\right)}{r - 1}$ , which resembles $S(n) = 3^n - 1$ . We can quickly see that $r = 3$ and $a = 2$ , so the $n^{\text{th}}$ term is $U(n) = 2 \cdot 3^{n - 1}$ with first term $a_1 = 2$ . Thus, we have

$\displaystyle \begin{aligned} \frac{S(2n) - S(n)}{U(2n) - U(n)} = \frac{\left(3^{2n} - 1\right) - \left(3^n - 1\right)}{\left(2 \cdot 3^{2n - 1}\right) - \left(2 \cdot 3^{n - 1}\right)} = \frac{3^n}{2 \cdot 3^{n - 1}} = \boxed{\dfrac 32} \end{aligned}$

From $S(1) =3-1 = 2$ , then the first term $U(1) = 2$ . $S(2) = U(1)+U(2) = 2 + 2r = 3^2-1=8$ , where $r$ is the common ratio. Then $\implies r = 3$ and $U(n) = 2(3^{n-1})$ . Therefore, we have:

$\begin{aligned} \frac {S(2n)-S(n)}{U(2n)-U(n)} & = \frac {3^{2n}-1 - 3^n + 1}{2(3^{2n-1})-2(3^{n-1})} \\ & = \frac {3^n\left(3^n -1\right)}{2\cdot 3^{n-1}\left(3^n - 1\right)} \\ & = \frac 32 = \boxed{1.5} \end{aligned}$