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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem

Find an approximation of $(0.99)^5$ using the first three terms of its expansion.

$\begin{array}{1 1}(A)\;951\\(B)\;0.951\\(C)\;0.0951\\(D)\;0.851\end{array} $

1 Answer

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  • $(1-x)^n=nC_0-nC_1x+nC_2x^2-.......+nC_r(-x)^r+.......+nC_n(-x)^n$
$(0.99)^5=(1-0.01)^5$
$\Rightarrow 1-5C_1\times (0.01)+5C_2\times (0.01)^2$......
$\Rightarrow 1-0.05+10\times 0.0001$
$\Rightarrow 1.001-0.05$
$\Rightarrow 0.951$
Hence (B) is the correct answer.
answered Jun 24, 2014 by sreemathi.v
 
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