The sum of the coefficients of x^{32} and x^{ | vedclass

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

Vedclass · Accepted Answer

$1260$

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(B) The general term in the expansion of $\left(x^4-\frac{1}{x^3}ight)^{15}$ is given by $T_{r+1} = \binom{15}{r} (x^4)^{15-r} (-x^{-3})^r = \binom{15}{r} (-1)^r x^{60-4r-3r} = \binom{15}{r} (-1)^r x^{60-7r}$.For the coefficient of $x^{32}$,we set $60-7r = 32$,which gives $7r = 28$,so $r = 4$.The coefficient is $\binom{15}{4} (-1)^4 = \frac{15 	imes 14 	imes 13 	imes 12}{4 	imes 3 	imes 2 	imes 1} = 1365$.For the coefficient of $x^{-31}$,we set $60-7r = -31$,which gives $7r = 91$,so $r = 13$.The coefficient is $\binom{15}{13} (-1)^{13} = \binom{15}{2} (-1) = -\frac{15 	imes 14}{2 	imes 1} = -105$.The sum of the coefficients is $1365 + (-105) = 1260$.

The sum of the coefficients of $x^{32}$ and $x^{-31}$ in the expansion of $\left(x^4-\frac{1}{x^3}\right)^{15}$ is:

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