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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Expand the expression: $(1-2x)^5$

$\begin{array}{1 1} 1-10x+40x^2-80x^3+90x^4-32x^5 \\1+10x-40x^2+80x^3-90x^4+32x^5 \\1-10x+40x^2-80x^3+80x^4-32x^5 \\ 1+10x-40x^2-80x^3-90x^4+16x^5\end{array} $

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  • $(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$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
Given $(1-2x)^5$:
$\Rightarrow (1-2x)^5 = \large \sum \limits_{k=0}^{5}\; $$^5 \large C$$_k \; a^{5-k}b^k $
$\qquad = \; ^5 \large C$$_0 \; 1^{5-0}(-2x)^0 +\; ^5 \large C$$_1 \; 1^{5-1}(-2x)^{1} + ^5 \large C$$_2 \; 1^{5-2}(-2x)^2 + ^5 \large C$$_3 \; 1^{5-3}(-2x)^3 + ^5 \large C$$_4 \; 1^{5-4}(-2x)^4 + ^5 \large C$$_5 \; 1^{5-5}(-2x)^5$
$\qquad = (1-2x)^5 = 1 - 5 (2x) + 10(4x^2) - 10(8x^3) + 5(16x^4) - 32(x^5)$
$\qquad = (1-2x)^5 = 1-10x+40x^2-80x^3+80x^4-32x^5$
answered Apr 9, 2014 by balaji.thirumalai
 

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