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Answered

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Evaluate:
Σ015 (3)
n-1
?

Evaluate Σ015 3 N1 class=

Sagot :

abidemiokin

The sum of the given geometric progression to the nearest hundredth is 7.50

Sum of a geometric progression

The given nth term is an explicit formula for a geometric sequence. The formula for calculating the sum of the given expression is;

Sum = a(1-r^n)/1-r

where

r is the common ratio. = 1/2

n is the number of terms = 10

a is the first term = 5

Substitute to have:

Sum = 5(1-(1/3)^10)/1-1/3

Sum = 5(1-1/59049)/2/3

Sum = (5 - 5/59049) * 3/2

sum = 7.50

Hence the sum of the given geometric progression to the nearest hundredth is 7.50

Learn more on sum of geometric sequence here: https://brainly.com/question/24643676

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