Consider the infinite geometric series ∑∞n=1−4(13)n−1

. in this image, the lower limit of the summation notation is "n = 1".

a. write the first four terms of the series.
b. does the series diverge or converge?
c. if the series has a sum, find the sum.

See below

Step-by-step explanation:

-4 (1/3)^(n-1)

Part A

For n = 1

-4(1/3)^(1 - 1)

-4(1/3)^0    Anything to the 0 power (except 0) is 1

-4 (1)

-4

n = 2

-4(1/3)^(2 - 1)

-4*(1/3)^1

-4/3

n = 3

- 4(1/3)^2

-4/9

n = 4

-4/(1/3)^3

-4 / 27

Part B

The series converges.

1/3 is between -1 <= 1/3 <= 1

Part C

Formula

Sum = a/(1 - r)

a = - 4

r = 1/3

Sum = -4/(1 - 1/3) = -4//2/3 = - 4/(0.666666666...) = -6

the answer is 62.8 cubic units