Term independent of x in {left( {2x - frac{1}{{2{x^2}}}} right)^{12}} is

English

Gujarati

Hindi

7.Binomial Theorem

easy

The term independent of $x$ in ${\left( {2x - \frac{1}{{2{x^2}}}} \right)^{12}}$is

A

$-7930$

B

$-495$

C

$495$

D

$7920$

Solution

(d) We have ${(x)^{12 – r}}{\left( {\frac{1}{{{x^2}}}} \right)^r} = {x^0}\,\, \Rightarrow {x^{12 – 3r}} = {x^0} \Rightarrow r = 4$

Hence the required term is $^{12}{C_4}{2^8}{\left( { – \frac{1}{2}} \right)^4} = 7920$.

Standard 11

Mathematics

Share

Similar Questions

In the expansion of ${(1 + 3x + 2{x^2})^6}$ the coefficient of ${x^{11}}$ is

hard

The coefficient of $x^{18}$ in the product $(1+ x)(1- x)^{10} (1+ x + x^2 )^9$ is

hard

(JEE MAIN-2019)

Let $\alpha$ be the constant term in the binomial expansion of $\left(\sqrt{ x }-\frac{6}{ x ^{\frac{3}{2}}}\right)^{ n }, n \leq 15$. If the sum of the coefficients of the remaining terms in the expansion is $649$ and the coefficient of $x^{-n}$ is $\lambda \alpha$, then $\lambda$ is equal to $……….$.

hard

(JEE MAIN-2023)

If in the expansion of ${(1 + x)^m}{(1 – x)^n}$, the coefficient of $x$ and ${x^2}$ are $3$ and $-6$ respectively, then m is

hard

(IIT-1999)

The positive value of $\lambda $ for which the co-efficient of $x^2$ in the expression ${x^2}{\left( {\sqrt x + \frac{\lambda }{{{x^2}}}} \right)^{10}}$ is $720$ is

hard

(JEE MAIN-2019)