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

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(D) The general term in the expansion of $(x^4-\frac{1}{x^3})^{15}$ is given by $T_{r+1} = {}^{15}C_r (x^4)^{15-r} (-\frac{1}{x^3})^r$.This simplifies to $T_{r+1} = {}^{15}C_r (-1)^r x^{60-4r} x^{-3r} = {}^{15}C_r (-1)^r x^{60-7r}$.To find the coefficient of $x^{18}$,we set the exponent $60-7r = 18$.$7r = 60 - 18 = 42$,which gives $r = 6$.The coefficient is ${}^{15}C_6 (-1)^6 = {}^{15}C_6 = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 5005$.

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 $x^{18}$ in the expansion of $(x^4-\frac{1}{x^3})^{15}$ is $...........$.

JEE MAIN 2023 All JEE MAIN

A
$5004$
B
$5003$
C
$5002$
D
$5005$

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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 $x^{18}$ in the expansion of $(x^4-\frac{1}{x^3})^{15}$ is $...........$.

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