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Question
Use binomial theorem to evaluate the following upto four places of decimal
`sqrt(99)`
Solution
`sqrt(99) = (99)^(1/2)`
= `(100 - 1)^(1/2)`
= `[100(1 - 1/100)]^(1/2)`
= `(100)^(1/2) (1 - 0.01)^(1/2)`
= `10[1 - 1/2(0.01) + (1/2(1/2 - 1))/(2!) (0.01)^2 - ......]`
= `10[1 - 1/2(0.01) + (1/2((-1)/2))/2 (0.01)^2 - ......]`
= `10[1 - 0.005 - (0.0001)/8 - ......]`
= 10 [1 – 0.005 – 0.0000125 – ……]
= 10 [0.9949875]
= 9.949875
= 9.9499
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