The sum of the series is: (c) lim n⇒∝ 1/3(1 - (1/4)^n)
How to evaluate the sum of the series?
The complete question is added as an attachment
The given series is a geometric series. So, we first calculate the common ratio (r) using:
r = T2/T1
From the series, we have:
T1 = 1/4 and T2 = 1/16
Substitute the known values in the above equation
r = (1/16)/(1/4)
So, the equation becomes
T = 1/16 / 1/4
Rewrite as product
T = 1/16 * 4
Evaluate the product
r = 1/4
The formula to calculate the sum of a geometric series of is:
Sn = a(1 - r^n)/(1 - r)
Where
a = 1/4 -- the first term
r = 1/4 --- the common ratio
Substitute the known values in the above equation
Sn = 1/4 * (1 - (1/4)^n)/(1 - 1/4)
Simplify the denominator
Sn = 1/4 * (1 - (1/4)^n)/(3/4)
Divide 1/4 by 3/4
Sn = 1/3 * (1 - (1/4)^n)
Take the limit to infinity of the series
Sn = lim n⇒∝ 1/3(1 - (1/4)^n)
We can conclude that, the sum of the series is: (c) lim n⇒∝ 1/3(1 - (1/4)^n)
Read more about geometric series at:
brainly.com/question/21087466
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