Question

4 + 12 + 36 + … ; n = 15

Answer 1

4 times 3=12
12 times 3=36
I think you want us to find the sum of the geometric sequence to n=15

the sum is
$S= \frac{a_1(1-r^{n})}{1-r}$
a1=first term=4
n=15
r=common ratio=3

$S= \frac{4(1-3^{15})}{1-3}$
$S= \frac{4(1-14348907)}{-2}$
S=-2(-14348906)
S=28697812

it adds to 28697812

Answer 2

I've attached a picture below of the formula for a finite geometric sequence. It includes an infinite geometric sequence formula for your reference, too.

$a_1 = 4$
$r(ratio)= 3$
$n=15$

Plug in the values. 

$S_{15} = 4*( \frac{1-3^{15}}{1-3} )$
$S_{15} = 4*( \frac{1-3^{15}}{-2} )$
$S_{15} = 4*( \frac{1-14348907}{-2} )$
$S_{15} = 4*( \frac{-14348906}{-2} )$
$S_{15} = 4*7174453$
S₁₅ = 28697812