AP ICET 2016 Paper

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Question : 136 of 200

Marks: +1, -0

The value of the term independent of x in the expansion of

(3x-\;\;{4}/{x^{2}})^{9}

is

$7×3^{7}×2^{8}$
$7×3^{8}×2^{7}$
$-7×3^{7}×2^{8}$
$-7×3^{8}×2^{7}$

Solution:

Let

t_{r+1}

be the term independent of x.

∴r={np}/{p+q}

Given expression :

(3x-4/x^2)^9

Here,

n=9,p=1,q=2

∴r={9×1}/{1+2}=3

∴{t}_{4}=^{9}{C}_{3}(3)^{9-3}(-4)^{3}=({9×8×7}/{3×2})(3^{6})(-4)^{3}

=7×3×4×3^{6}×4^{3}

=7×3^{7}×2^{8}

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