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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem

Find the $13th$ term in the expansion of $\large ($$9x$$-\large \frac{1}{3\sqrt x})^{\normalsize 18}$$, x\neq 0$

$\begin{array}{1 1} 18564 \\ 19376 \\ 21364 \\ 15864 \end{array} $

1 Answer

Toolbox:
  • $(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$$_r\;a^{n–r}b^r$.
Given $\large ($$9x$$-\large \frac{1}{3\sqrt x})^{\normalsize 18}$$, a =9x, b = \large\frac{-1}{3\sqrt x}$$, n = 18, r = 12$
$\Rightarrow T_{12+1} = ^{18} \large C$$_{12}\;(9x)^{18-12} \large(\frac{-1}{3\sqrt x})$$^{\normalsize 12}$
$\Rightarrow T_{13} = \; ^{18}\large C$$_{12}$$ \;9^6 x^6 \large\frac{1}{(\sqrt x)^{12}}$$ \large\frac{1}{3^{12}}$
$\Rightarrow T_{13} = 18564 $
answered Apr 10, 2014 by balaji.thirumalai
edited Apr 10, 2014 by balaji.thirumalai
 

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