The constant term in the binomial expansion of (3x^{2}- \frac{1}{4x})^{6} is

AP ICET 2015 Paper

Question : 130 of 200

Marks: +1, -0

The constant term in the binomial expansion of

(3x^{2}-\;\;{1}/{4x})^{6}

Solution:

(3x^{2}-\;\;{1}/{4x})^{6}

Here, n = 6, p = 2, q = 1
If Tr + 1 is the constant term, then

r={np}/{p+q}={6×2}/{2+1}=4

∴ Constant term is

T_5=6C_4(3)^{6-4}(-1/4)^4

=({6×5}/{2})×(3)^{2}({1}/{4})^{4}={15×9}/{256}={135}/{256}

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