(C) The general term $T_{r+1}$ in the expansion of $(x^4 - x^{-3})^{15}$ is given by:$T_{r+1} = ^{15}C_r (x^4)^{15-r} (-x^{-3})^r$$T_{r+1} = ^{15}C_r

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(C) The general term $T_{r+1}$ in the expansion of $(x^4 - x^{-3})^{15}$ is given by:$T_{r+1} = ^{15}C_r (x^4)^{15-r} (-x^{-3})^r$$T_{r+1} = ^{15}C_r (-1)^r x^{60-4r} x^{-3r}$$T_{r+1} = ^{15}C_r (-1)^r x^{60-7r}$To find the coefficient of $x^{32}$,we set the exponent of $x$ equal to $32$:$60 - 7r = 32$$7r = 28$$r = 4$Substituting $r = 4$ into the expression,the coefficient is $^{15}C_4 (-1)^4 = ^{15}C_4$.

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

Easy

The coefficient of $x^{32}$ in the expansion of $(x^4 - \frac{1}{x^3})^{15}$ is

A
$^{15}C_5$
B
$^{15}C_6$
C
$^{15}C_4$
D
$^{15}C_7$

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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

The term independent of $x$ in ${\left( {\sqrt x - \frac{2}{x}} \right)^{18}}$ is

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If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$,then $\sum \limits_{r=1}^{10} r^3\left(\frac{a_r}{a_{r-1}}\right)^2$ is equal to

DifficultJEE MAIN 2023

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The second,third and fourth terms in the binomial expansion $(x+a)^n$ are $240, 720$ and $1080$ respectively. Find $x, a$ and $n$.

Difficult

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If the coefficients of $r$-th and $(r+1)$-th terms in the expansion of $(1+x)^{24}$ are in the ratio $12:13$,then $r$ is the root of the quadratic equation

EasyAP EAMCET 2019

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Find the coefficient of $a^{5} b^{7}$ in $(a-2 b)^{12}$.

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The coefficient of $x^{32}$ in the expansion of $(x^4 - \frac{1}{x^3})^{15}$ is

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The term independent of $x$ in ${\left( {\sqrt x - \frac{2}{x}} \right)^{18}}$ is

If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$,then $\sum \limits_{r=1}^{10} r^3\left(\frac{a_r}{a_{r-1}}\right)^2$ is equal to

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