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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Expand the expression: $(2x-3)^6$

$\begin{array}{1 1} 64x^6 - 576 x^5 + 2160 x^4 - 4320 x^3 + 4860 x^2 - 2916 x + 729 \\ 64x^6 + 576 x^5 - 2160 x^4 + 4320 x^3 - 4860 x^2 + 2916 x + 729 \\ 64x^6 + 576 x^5 - 2160 x^4 + 4320 x^3 - 4860 x^2 + 2916 x - 729 \\ 64x^6 - 576 x^5 + 2160 x^4 - 4320 x^3 + 4860 x^2 - 2916 x - 729 \end{array} $

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1 Answer

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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$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
Given $(2x-3)^6 \rightarrow a = 2x,\; b = -3$:
$\Rightarrow (2x-3)^6 = \large \sum \limits_{k=0}^{6}\; $$^6 \large C$$_k \; (2x)^{6-k}(-3)^k $
$\qquad = ^6 \large C$$_0 \; (2x)^{6-0}(-3)^0 + ^6 \large C$$_1 \; (2x)^{6-1}(-3)^1 + ^6 \large C$$_2 \; (2x)^{6-2}(-3)^2 + ^6 \large C$$_3 \; (2x)^{6-3}(-3)^3 + ^6 \large C$$_4 \; (2x)^{6-4}(-3)^4 + ^6 \large C$$_5 \; (2x)^{6-5}(-3)^5 + ^6 \large C$$_6 \; (2x)^{6-6}(-3)^6 $
$\qquad = 64x^6 - 6(32x^5)(3) + 15(16x^4)9 - 20(8x^3)27 + 15(4x^2)(81) - 6(2x)(243) + 729$
$\qquad = 64x^6 - 576 x^5 + 2160 x^4 - 4320 x^3 + 4860 x^2 - 2916 x + 729$
answered Apr 9, 2014 by balaji.thirumalai
edited Apr 9, 2014 by balaji.thirumalai
 

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