The coefficient of x^{18} in the expansion of | vedclass

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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

Vedclass · Accepted Answer

$5005$

Vedclass · Answer

(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 ${}^{15}C_6 (-1)^6 = {}^{15}C_6 = \frac{15 	imes 14 	imes 13 	imes 12 	imes 11 	imes 10}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} = 5005$.

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

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