What is the sum of the n terms of the series 4 + 12 + 36 + 108 + . . . ?

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asked Sep 10, 2017 11.4k views

1 Answer

HenryZhang · Answer 1 · 2017-09-15T08:01:31+0000

Sum of n terms of a geometric series is
Sn = a1 (r^n-1)/(r-1)
a1 = 1st term = 4
r = common ratio = 12/4 = 3
so, the sum is
Sn = 4 (3^n-1)/(3-1) = 2(3^n-1)

What is the sum of the n terms of the series 4 + 12 + 36 + 108 + . . . ?

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