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Question
Answer the following:
Find `sum_("r" = 1)^"n" (2/3)^"r"`
Solution
`sum_("r" = 1)^"n" (2/3)^"r" = 2/3 + (2/3)^2 + (2/3)^3 + ... + (2/3)^"n"`
The terms `2/3, (2/3)^2, (2/3)^3` are in G.P.
∴ a = `2/3`, r = `2/3`
∴ `sum_("r" = 1)^"n" (2/3)^"r" = (2/3[1 - (2/3)^"n"])/(1 - 2/3)`
∴ `sum_("r" = 1)^"n" (2/3)^"r" = 2[1 - (2/3)^"n"]`
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