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Question
Find : `sum_("r" = 1)^oo 4(0.5)^"r"`
Solution
`sum_("r" = 1)^oo 4(0.5)^"r" = 4 sum_("r" = 1)^oo (0.5)^"r"`
= 4[0.5 + (0.5)2 + (0.5)3 + ...]
= `4[5/10 + (5/10)^2 + (5/10)^3 + ...]`
= `4[1/2 + (1/2)^2 + (1/2)^3 + ....]` ...(1)
The terms `1/2, (1/2)^2, (1/2)^3` ... form a 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
∴ from (1),
`sum_("r" = 1)^oo 4(0.5)^"r"` = 4 x 1 = 4.
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