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If $x=\large\frac{1}{3}$, the greatest term in the expansion of $(1+4x)^8$ is ?

$\begin{array}{1 1} 3rd\;term \\ 4th\;term \\ 5th\;term \\ 6th\;term \end{array}$

Can you answer this question?
 
 

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General term $T_{r+1}$ in the expansion $(1+4x)^8$ = $^8C_r.4^r.x^r$
Let $(r)^{th}$ term be the greatest term.
Then $\large\frac{T_{r+1}}{T_r}$$\leq 1$
$\Rightarrow\:\large\frac{^8C_r.4^r.x^r}{^8C_{r-1}.4^{r-1}.x^{r-1}}$$\leq 1$
$\Rightarrow\:\large\frac{(9-r).4x}{r}$$\leq 1$
$\Rightarrow\:12-\large\frac{4r}{3}$$\leq r$
$\Rightarrow\:r\geq \large\frac{36}{7}$ But $r\in Z$ and $ r\leq 8$
$\Rightarrow\:r=6$
$\therefore$ The greatest term is 6th term.

 

answered Sep 17, 2013 by rvidyagovindarajan_1
edited Dec 23, 2013 by meenakshi.p
 

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