Answer
$3\times\frac{(3^n-1)}{2}$
Work Step by Step
We know that,
$1 + r + r^2 + ··· + r^n = \frac{r^{n+1}− 1}{r − 1}$
$3 + 3^2 + 3^ 3 + ··· + 3 ^n$
=$3\times(1+3 + 3^2 + ··· + 3 ^{n-1})$
=$3\times\frac{3^{(n-1)+1}-1}{2}$
=$\frac{3\times(3^n-1)}{2}$ or $\frac{3^{n+1}-3}{2}$
