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Question
Find `sum_("r" = 0)^oo (-8)(-1/2)^"r"`
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Solution
`sum_("r" = 0)^oo (-8)(-1/2)^"r" = -8 sum_("r" = 1)^oo (-1/2)^"r"`
= `-8[(-1/2) + (-1/2)^2 + (-1/2)^3 + ...]` ...(1)
The terms `(-1/2), (-1/2)^2, (-1/2)^3 ...` are in G.P. with a = `-1/2`, r = `-1/2`.
Since |r| = `|-1/2| = 1/2 < 1`, the sum to infinity of this G.P. exist and
S = `"a"/(1 - "r")`
= `((-1/2))/(1 - (-1/2))`
= `-1/2 xx 2/3`
= `(-1)/3`
∴ from (1),
`sum_("r" = 0)^oo (-8)(-1/2)^"r" = -8(-1/3) = 8/3`
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