Consider the infinite geometric series ∑∞n=1−4(13)n−1
2 answers:
Answer:
See below
Step-by-step explanation:
-4 (1/3)^(n-1)
Part A
<em><u>For n = 1</u></em>
-4(1/3)^(1 - 1)
-4(1/3)^0 Anything to the 0 power (except 0) is 1
-4 (1)
-4
<em><u>n = 2</u></em>
-4(1/3)^(2 - 1)
-4*(1/3)^1
-4/3
<em><u>n = 3</u></em>
- 4(1/3)^2
-4/9
<em><u>n = 4</u></em>
-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
Answer:
-4 (1/3)^(n-1)
Step-by-step explanation:
This is correct. Connexuss.
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Answer:
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Step-by-step explanation:
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PEMDAS
Parentheses first
12+[(10)]+(6)]
Then add
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Answer:
what?Can you write better?
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