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प्रश्न
Find : `sum_("r" = 1)^oo (-1/3)^"r"`
उत्तर
`sum_("r" = 1)^oo (-1/3)^"r"`
= `(-1/3) + (-1/3)^2 + (-1/3)^3 + ...`
The terms `(-1/3), (-1/3)^2, (-1/3)^3` are in G.P.
∴ a = `-1/3`, r = `-1/3`
Since, |r| = `|-1/3| < 1`
∴ sum to infinity exists.
∴ `sum_("r" = 1)^oo (-1/3)^"r" = (-1/3)/(1 - (-1/3))`
= `(-1/3)/(4/3)`
= `-1/4`
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