Question

find the value of √99 up to four decimal placse using binomial theorem​

Answer 1

value of √(99) to four places is 9.9498

we have to find the value of √(99) upto four decimal places using binomial theorem.

concept : (1 - x)ⁿ = 1 - nx + n(n -1)x²/2! - n(n - 1)(n -2)x³/3! + n(n -1)(n-2)(n-3)x⁴/4! ..... ∞

here, √(99) =√(100 - 1)

= {100(1 - 1/100)}½

= 10(1 - 0.01)½

now, (1 - 0.01)½ seems (1 -x)ⁿ

so, (1 - 0.01)½ = 1 - 1/2 × (0.01) + (1/2)(1/2-1)(0.01)²/2 - (1/2)(1/2-1)(1/2-2)(0.01)³/3! + .....

= 1 - 0.005 + (1/2)(-1/2)(0.0001)/2 - (1/2)(-1/2)(-3/2)(0.000001)/6 + ...

= 1 - 0.005 - 0.0000125 - ....

= 0.9949875

so, 10(1 - 0.01)½ = 10 × 0.9949875

= 9.949875

value of √(99) to four places is 9.9498

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