Answer
$a_n=2 (4^{n-1})$
$a_{15}=536870912$
$S_{15}=715827882$
Work Step by Step
The general formula for the nth term of a geometric series is given by $a_n= a_1r^{n-1}$ ...(1)
The ratio of the successive terms is $4$.
Equation (1) gives: $a_n= a_1r^{n-1}=2 \times 4^{n-1}$
Plugging in $n =15$, we have
$a_{15}=2 \times 4^{15-1}=536870912$
We know that $S_{n}=\dfrac{a_1(1-r^n)}{1-r}$
Now, $S_{15}=\dfrac{2(1-4^{15})}{1-4}=715827882$
