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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Using Binomial Theorem, evaluate $99^5$

$\begin{array}{1 1} 9509900499 \\ 2568938 \\ 3198754 \\ 95279531 \end{array} $

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Toolbox:
  • Express the given number as the sum or difference of two numbers whose powers are easier to evaluate. Then use binomial theorem as follows:
  • $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
$99 = 100 - 1 \rightarrow 99^5 = (100-1)^5$. We can now use Binomial Theorem to evaluate this.
$(100-1)^5 = ^5 \large C$$_0\;100^{5-1}1^0 -.^5 \large C$$_1\;100^{5-1}1^1 +.^5 \large C$$_2\;100^{5-2}1^2 -.^5 \large C$$_3\;100^{5-3}1^3 + ^5 \large C$$_5\;100^{5-5}1^5 $
$\qquad = 100^5 - 5(100^4) + 10(100^3) - 10(100^2) + 5(100) - 1$
$\qquad = 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1$
$\qquad = 9509900499$
answered Apr 10, 2014 by balaji.thirumalai
 

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