The coefficient of {x^{39}} in the expansion | vedclass

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Question

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

Vedclass · Accepted Answer

$-455$

Vedclass · Answer

(A) The general term in the expansion of ${\left( {{x^4} - \frac{1}{{{x^3}}}} ight)^{15}}$ is given by:${T_{r + 1}} = {}^{15}{C_r} {({x^4})^{15 - r}} {\left( - \frac{1}{{{x^3}}} ight)^r}$${T_{r + 1}} = {}^{15}{C_r} {(-1)^r} {x^{60 - 4r}} {x^{-3r}}$${T_{r + 1}} = {(-1)^r} {}^{15}{C_r} {x^{60 - 7r}}$To find the coefficient of ${x^{39}}$,we set the exponent of $x$ equal to $39$:$60 - 7r = 39$$7r = 21$$r = 3$Substituting $r = 3$ into the general term expression:Coefficient $= {(-1)^3} {}^{15}{C_3}$$= -1 	imes \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1}$$= -1 	imes 5 	imes 7 	imes 13 = -455$Thus,the required coefficient is $-455$.

The coefficient of ${x^{39}}$ in the expansion of ${\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}$ is

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