(B) The general term in the expansion of $(x - \frac{1}{x})^7$ is given by $T_{r+1} = ^7C_r (x)^{7-r} (-x^{-1})^r$.This simplifies to $T_{r+1} = ^7C_r

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(B) The general term in the expansion of $(x - \frac{1}{x})^7$ is given by $T_{r+1} = ^7C_r (x)^{7-r} (-x^{-1})^r$.This simplifies to $T_{r+1} = ^7C_r (-1)^r x^{7-r-r} = ^7C_r (-1)^r x^{7-2r}$.To find the coefficient of $x^3$,we set the exponent $7 - 2r = 3$.$7 - 3 = 2r$ $\Rightarrow 2r = 4$ $\Rightarrow r = 2$.Substituting $r = 2$ into the expression,the coefficient is $^7C_2 (-1)^2 = \frac{7 \times 6}{2 \times 1} \times 1 = 21$.

Class 11 Mathematics Binomial Theorem General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

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The coefficient of $x^3$ in the expansion of $(x - \frac{1}{x})^7$ is

A
$14$
B
$21$
C
$28$
D
$35$

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Binomial Theorem

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General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

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Find the middle terms in the expansions of $\left(3-\frac{x^{3}}{6}\right)^{7}$.

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DifficultJEE MAIN 2021

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The index of the power of $x$ occurring in the $5^{\text{th}}$ term from the end in the expansion of $\left(\frac{x^3}{2} - \frac{2}{x^2}\right)^{12}$ is

EasyAP EAMCET 2020

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The coefficient of $x^3$ in the expansion of $(x - \frac{1}{x})^7$ is

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