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3 years ago

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

Mathematics

2 answers:

vodomira [7]3 years ago

3 0

Answer:

See below

Step-by-step explanation:

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

Part A

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

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

-4 (1)

-4

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

-4*(1/3)^1

-4/3

- 4(1/3)^2

-4/9

-4/(1/3)^3

-4 / 27

Part B

The series converges.

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

Part C

<em><u>Formula</u></em>

Sum = a/(1 - r)

a = - 4

r = 1/3

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

Basile [38]3 years ago

3 0

Answer:

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

Step-by-step explanation:

This is correct. Connexuss.

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