Q.For the following G.P., find $S_n$: $3, 6, 12, 24, \ldots$

For a G.P. $a, ar, ar^2, \ldots, ar^{n-1}$ with $r\ne1$, the sum of the first $n$ terms is $S_n=\dfrac{a(r^n-1)}{r-1}$, equivalently written $\dfrac{a(1-r^n)}{1-r}$ when $r<1$; this is derived by multiplying $S_n$ by $r$ and subtracting the two series, so every middle term cancels and only a first and a last term survive. When $r=1$, every term equals $a$, so $S_n=na$. A related identity, $S_n-S_{n-1}=t_n$, is useful for recovering the $n$th term when only a formula for $S_n$ is given, and can also be used to prove that a sequence given by its $S_n$ is a G.P. Repeating-digit sums such as $3+33+333+\cdots$ or $0.4+0.44+0.444+\cdots$ become genuine G.P. sums once factored and each block like $99\ldots9$ is rewritten as $10^k-1$.