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  4. Using binomial theorem, the value of (0.999)3 correct to 3 decimal places is

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Q. Using binomial theorem, the value of $(0.999)^{3}$ correct to $3$ decimal places is

WBJEEWBJEE 2009Binomial Theorem

Solution:

$\left(0.999\right)^{3}=\left(1-0.001\right)^{3}$

$=\,{}^{3}C_{0}-\,{}^{3}C_{1}\left(0.001\right)+\,{}^{3}C_{2}\left(0.001\right)^{2}-\,{}^{3}C_{3}\left(0.001\right)^{3}$

$=1-0.003+3\left(0.000001\right)-\left(0.000000001\right)$

$=0.997$