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VITEEE 2010 Paper

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Question : 83 of 120

Marks: +1, -0

The numerically greatest term in the expansion of

(3 - 5x)^{11}

x=\frac{1}{5},

is

$55\times3^{9}$
$55\times3^{6}$
$45\times3^{9}$
$45\times3^{6}$

Solution:

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

= 3^{11}\left(1-\frac{5}{3}\cdot\frac{1}{5}\right)^{11}

{\because\; x = \frac{1}{5}}

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

Now,

r=\frac{|x|(n+1)}{|x|+1}=\frac{\left|-\frac{1}{3}\right|(11+1)}{\left|-\frac{1}{3}\right|+1}=\frac{4}{4/3}

⇒ r = 3

Therefore, 3rd

(T_{3})

and (3+1) = 4th

(T_{4})

terms are numerically greatest in the expansion of

(3-5x)^{11}.

So, greatest term =

T_{3}

=3^{11}\left|\binom{11}{2}(1)^{9}\left(-\frac{1}{3}\right)^{2}\right|

=3^{11}\left|\frac{11\times10}{1\cdot2\cdot9}\right|

=55\times3^{9}

and

T_{4}=3^{11}\left|\binom{11}{3}(1)^{8}\left(-\frac{1}{3}\right)^{3}\right|

=3^{11}\left|\frac{11\times10\times9}{1\cdot2\cdot3}\cdot\left(-\frac{1}{27}\right)\right|

=55\times3^{9}

∴ Greatest term(numerically)

= T_{3} = T_{4} =55\times3^{9}

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