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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
0 votes

Expand the expression:$(99)^5$

$\begin{array}{1 1}(A)\;9509900498\\(B)\;9509900499\\(C)\;9509900497\\(D)\;\text{None of these}\end{array} $

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1 Answer

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Toolbox:
  • $(a-b)^n=nC_0a^n-nC_1a^{n-1}b+nC_2a^{n-2}b^2-....+(-)^nnC_ra^{n-r}b^r+.......+nC_n(-b)^n$
$(100-1)^5=5C_0(100)^5+5C_1(100)^4(-1)+5C_2(100)^3(-1)^2+5C_3(100)^2(-1)^3+5C_4(100).(-1)^4+(-1)^5$
$\Rightarrow 10000000000+(-5)\times 100000000+10\times 1000000-10\times 10000+5\times 100-1$
$\Rightarrow 10000000000-500000000+10000000-100000+500-1$
$\Rightarrow 9509900499$
Hence (B) is the correct answer.
answered Jun 23, 2014 by sreemathi.v
 
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