What is the 4th partial sum for the geometric sequence where a1=18 and r = 2 a. -18
b. 288
c. 270
d. 152

The 4th partial sum of the sequence is the sum of the first 4 terms:

S₄ = a(1) + a(2) + a(3) + a(4)

Because the sequence is geometric, any term can be written in terms of the previous one:

a(n) = r a(n - 1)

and, more importanly, in terms of the first term a(1), so that

a(2) = r a(1)

a(3) = r a(2) = r ² a(1)

a(4) = r a(3) = r ³ a(1)

Then the 4th partial sum is

S₄ = a(1) + r a(1) + r ² a(1) + r ³ a(1)

S₄ = a(1) (1 + r + r ² + r ³)

With a(1) = 18 and r = 2, we have

S₄ = 18 (1 + 2 + 4 + 8) = 270

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8

the answer is d on edge

step-by-step explanation: