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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.
a. To find the first four terms of the series, we substitute n=1, 2, 3, 4 into the formula:
n = 1: -4(1/3)^0 = -4
n = 2: -4(1/3)^1 = -4/3
n = 3: -4(1/3)^2 = -4/9
n = 4: -4(1/3)^3 = -4/27
Therefore, the first four terms of the series are -4, -4/3, -4/9, -4/27.
b. To determine if the series converges or diverges, we must first find the common ratio of the geometric series. In this case, the common ratio is -1/3. Since the absolute value of the common ratio is less than 1 (|r| < 1), the series converges.
c. To find the sum of the series, we can use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
where a is the first term and r is the common ratio.
Plugging in the values, we get:
S = -4 / (1 - (-1/3))
S = -4 / (4/3)
S = -3
Therefore, the sum of the series is -3.