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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Write the general term in the expansion of $ (x^2 - yx)^{12},\; x\neq 0$

$\begin{array}{1 1} -\;^{12} \large C_r\;x^{24-r} y^r \\-\;^{12} \large C_r\;x^{24-2r} y^r \\-\;^{12} \large C_r\;x^{12-r} y^r \\ -\;^{12} \large C_r\;x^{24-2r} y^{2r} \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$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
  • Looking at the pattern of the successive terms we can say that the $(r + 1)\text{th}$ term $T_{r+1} = ^n\large C$$_r\;a^{n–r}b^r$. This is also called the general term of the expansion.
$T_{r+1} = ^n\large C$$_r\;a^{n–r}b^r$.
Given $(x^2-yx)^{12} \rightarrow a = x^2, b = (-yx), n = 12$
$\Rightarrow T_{r+1} = ^{12} \large C$$_r\;(x^2)^{12-r} (-yx)^r = -\;^{12} \large C$$_r\;x^{24-r} y^r $
answered Apr 10, 2014 by balaji.thirumalai
 

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