In the expansion of {left( 2{x^2} - frac{1}{x | vedclass

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(B) The general term in the expansion of ${\left( 2{x^2} - \frac{1}{x} \right)^{12}}$ is given by: ${T_{r + 1}} = {}^{12}{C_r} \cdot {(2{x^2})^{12 - r

Vedclass · Accepted Answer

$9$

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(B) The general term in the expansion of ${\left( 2{x^2} - \frac{1}{x} \right)^{12}}$ is given by: ${T_{r + 1}} = {}^{12}{C_r} \cdot {(2{x^2})^{12 - r}} \cdot {\left( -\frac{1}{x} \right)^r}$ ${T_{r + 1}} = {}^{12}{C_r} \cdot {2^{12 - r}} \cdot {x^{24 - 2r}} \cdot {( - 1)^r} \cdot {x^{-r}}$ ${T_{r + 1}} = {}^{12}{C_r} \cdot {2^{12 - r}} \cdot {( - 1)^r} \cdot {x^{24 - 3r}}$ For the term to be independent of $x$,the exponent of $x$ must be $0$: $24 - 3r = 0$ $3r = 24$ $r = 8$ Since the term is ${T_{r + 1}}$,we have ${T_{8 + 1}} = {T_9}$,which is the $9^{th}$ term.

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

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