In expansion of {left( {2{x^2} - frac{1}{x}} right)^{12}} , term independent of x is

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7.Binomial Theorem

easy

In the expansion of ${\left( {2{x^2} - \frac{1}{x}} \right)^{12}}$, the term independent of x is

A

$10^{th}$

B

$9^{th}$

C

$8^{th}$

D

$7^{th}$

Solution

(b) ${T_{r + 1}} = {}^{12}{C_r}{(2{x^2})^{12 – r}}{( – 1)^r}{\left( {\frac{1}{x}} \right)^r}$

For term independent of $x$,

$24 – 3r = 0 \Rightarrow r = 8$ So, $9^{th}$ term independent of $x$.

Standard 11

Mathematics

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