Sum of Infinite Geometric series, alternate a values
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Sum of Infinite Geometric series, alternate a values

[From: ] [author: ] [Date: 11-12-16] [Hit: ]
........
If a is constant we have a/(1-r). But if a alternates is there any way I can solve this? The equation I'm trying to solve is:
(1-r)[3+2r+3r^2+2r^3+3r^4+...] where a alternates between 3 and 2.
I appreciate the help.

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3 + 2r + 3r^2 + 2r^3 + ... = (3 + 3r^2 + ... ) + (2r + 2r^3 + ...) = 3/(1 - r^2)) + 2r/(1 - r^2) = (3 + 2r)/(1 - r^2)
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