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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Using Binomial Theorem, evaluate $102^6$

$\begin{array}{1 1}884736 \\ 1124864 \\ 824864 \\ 1124736\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$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
$102 = 100+2 \rightarrow 102^5 = (100+2)^5$. We can now use Binomial Theorem to evaluate this.
$(100+2)^5= ^5 \large C$$_0\;100^{5-0}2^0 + ^5 \large C$$_1\;100^{5-1}2^1 + ^5 \large C$$_2\;100^{5-2}2^2 + ^5 \large C$$_3\;100^{5-3}2^3 + ^5 \large C$$_4\;100^{5-4}2^4 + ^5 \large C$$_5\;100^{5-5}2^5$
$\qquad = 100^5 + 5 (100^4) (2) + 10 (100^3) (2^2) + 10 (100^2) (2^3) + 5 (100) (2^4) + 2^5$
$\qquad = 10000000000 + 1000000000 + 40000000 + 800000+8000+32$
$\qquad = 11040808032$
answered Apr 10, 2014 by balaji.thirumalai
edited Apr 10, 2014 by balaji.thirumalai
 

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