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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find the coefficient of $a^5b^7$ in $ (a-2b)^{12}$

$\begin{array}{1 1} -101376 \\ 69341 \\ 45623\\ 62013\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.
In the expression $(a-2b)^{12}$, assume that $a^5b^7$ occurs in the $(r+1)\text{th}$ term.
$\Rightarrow T_{r+1} = \; ^{12} \large C$$_r\;a^{12–r}(-2b)^r$.
Comparing the indices of $a$ and $b$, we can see that $12 - r = 5 \rightarrow r = 7$
$\Rightarrow$ Coefficient of $a^5b^7 = \;^{12} \large C$$_7 \; a^{12-7}(-2b)^{7} $
$\Rightarrow$ Coefficient of $a^5b^7 =\;^{12}\large C$$_7\;(-2)^7$
$\qquad \qquad = 792 \times -128 = -101376$
answered Apr 10, 2014 by balaji.thirumalai
 

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