4 + 12 + 36 + … ; n = 15

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
[tex]S= \frac{a_1(1-r^{n})}{1-r} [/tex]
a1=first term=4
n=15
r=common ratio=3

[tex]S= \frac{4(1-3^{15})}{1-3} [/tex]
[tex]S= \frac{4(1-14348907)}{-2} [/tex]
S=-2(-14348906)
S=28697812

it adds to 28697812

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Аноним Аноним · Answer 2 · 2017-05-11T01:45:07+02:00

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.

[tex]a_1 = 4[/tex]
[tex]r(ratio)= 3 [/tex]
[tex]n=15[/tex]

Plug in the values.

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