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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find the $4th$ term in the expansion of $(x - 2y)^{12}$.

$\begin{array}{1 1}-1760 x^9y^3 \\ 1760 x^9y^3 \\1760 x^6y^3 \\ -1760 x^6y^3\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$$_r\;a^{n–r}b^r$.
Given $(x – 2y)^{12} \rightarrow a = x, b = -2y, n = 12, r = 3$
$\Rightarrow T_{3+1} = ^{12} \large C$$_3\;(x)^{12-3} (-2y)^3$
$\Rightarrow T_4 = -\; ^{12}\large C$$_3\; x^9 2^3 y^3 = -1760 x^9y^3$
answered Apr 10, 2014 by balaji.thirumalai
 

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