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3 votes
15 points please help

15 points please help-example-1
User BenvolioT
by
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2 Answers

7 votes

Hello from MrBillDoesMath!

Answer:

The second choice, a1 (1-r^n)/(1-r)

Discussion:

Let

S= a1 + a1*r + a1*r^2 +......+a1* r^(n-1) (*)

be a geometric series with n terms

Multiply both sides by "r":

Sr = a1*r + a1*r^2 +... + a1*r^(n-1) + a1*r^(n) (**)

Subtracting (*) from (**) gives

S-Sr = a1 - a1*r^(n)

= a1 ( 1 - r^(n))

As S=Sr = S (1-r)

S (1-r) = a1 ( 1 - r^(n))

Divide both sides by (1-r)

S = a1 ( 1 - r^(n)) /(1-r)

Thank you,

MrB

User Yanick Girouard
by
5.5k points
1 vote

Answer:

Sn = a1 (1-r^n)/ (1-r)

Explanation:

The formula for the sum of a geometric sequence summations is given by

Sn = a1 (1-r^n)/ (1-r)

where a1 is the first term, r is the common ratio and n is the term number we are summing up to

User Ryu Kent
by
5.3k points