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

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

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
$^{15}C_4$
B
$^{15}C_3$
C
$^{15}C_2$
D
$^{15}C_5$

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

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