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Find numerically the greatest term in the expansion of $(3-5 x)^{11}$ when $x=\frac{1}{5}$ .
Option: 1
$44 \times 3^{8}$

Option: 2
$55 \times 3^{9}$

Option: 3
$55 \times 3^{10}$

Option: 4
$44\times 3^{10}$

Answers (1)

best_answer

Since $(3-5 x)^{11}=3^{11}\left(1-\frac{5 x}{3}\right)^{11}$

Now in the expansion of $\left(1-\frac{5 x}{3}\right)^{11}$ , we have

$\frac{T_{r+1}}{T_{r}}=\frac{(11-r+1)}{r}\left|-\frac{5 x}{3}\right|$

$=\left(\frac{12-r}{r}\right)\left|-\frac{5}{3} \times \frac{1}{5}\right|$ $\because (x=\frac{1}{5}) \\$

$=\left(\frac{12-r}{r}\right)\left(\frac{1}{3}\right)\\$

$=\left(\frac{12-r}{3 r}\right) \\$

$\therefore \quad \frac{T_{r+1}}{T_{r}} \geq 1 \Rightarrow \frac{12-r}{3 r} \geq 1 \\$

$\Rightarrow \quad 4 r \leq 12 \\$

$\Rightarrow \quad r \leq 3 \therefore r=2,3$

so, the greatest terms are $\mathrm{T}_{2+1}$ and $\mathrm{T}_{3+1}$ .

$\therefore$ Greatest terms (where $r=2)=3^{11}\left|T_{2+1}\right|$

$= 3^{11}\left|{ }^{11} C_{2}\left(-\frac{5}{3} x\right)^{2}\right| \\$

$= 3^{11}\left|{ }^{11} C_{2}\left(-\frac{5}{3} \times \frac{1}{5}\right)^{2}\right| \quad\left( x=\frac{1}{5}\right) \\$

$= 3^{11}\left|\frac{11.10}{1.2} \times \frac{1}{9}\right|=55 \times 3^{9}$

and greatest term (where $r=3)=3^{11}\left|T_{3+1}\right|$

$= 3^{11}\left|{ }^{11} C_{3}\left(-\frac{5}{3} x\right)^{3}\right| \\$

$= 3^{11}\left|{ }^{11} C_{3}\left(-\frac{5}{3} \times \frac{1}{5}\right)^{3}\right| \\$

$= 3^{11}\left|\frac{11.10 .9}{1.2 .3} \times \frac{-1}{27}\right|=55 \times 3^{9}$

From above we say that the values of both greatest terms are equal.

Alternative Method (Short Cut Method) :

Since $(3-5 x)^{11}=3^{11}\left(1-\frac{5 x}{3}\right)^{11}=3^{11}\left(1-\frac{1}{3}\right)^{11} \quad\left( x=\frac{1}{5}\right)$

Now, calculate $m=\frac{|x|(n+1)}{(|x|+1)} \quad\left(-\frac{1}{3}<0\right)$

$=\frac{\left(\left|-\frac{1}{3}\right|\right)(11+1)}{\left(\left|-\frac{1}{3}\right|+1\right)} \\$

$=3$

$\therefore$ The greatest terms in the expansion are $\mathrm{T}_{3}\,\, and\,\, \mathrm{T}_{4}$

$\therefore$ Greatest term (when $r=2)=3^{11}\left|T_{2+1}\right|$

$=3^{11}\left|{ }^{11} C_{2}\left(-\frac{1}{3}\right)^{2}\right|=3^{11}\left|\frac{11.10}{1.2} \times \frac{1}{9}\right|=55 \times 3^{9}$

and greatest term (when $r=3)=3^{11}\left|T_{3+1}\right|$

$= 3^{11}\left|{ }^{11} C_{3}\left(-\frac{1}{3}\right)^{3}\right| \\$

$= 3^{11}\left|\frac{11.10 \times 9}{1.2 .3} \times \frac{-1}{27}\right|=55 \times 3^{9}$

From above we say that the values of both greatest terms are equal.

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Riya

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