The term independent of y in the expansion of | vedclass

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(D) The general term $T_{r+1}$ in the expansion of $(y^{-1/6} - y^{1/3})^9$ is given by:$T_{r+1} = ^9C_r (y^{-1/6})^{9-r} (-y^{1/3})^r$$T_{r+1} = ^9C_

Vedclass · Accepted Answer

$-84$

Vedclass · Answer

(D) The general term $T_{r+1}$ in the expansion of $(y^{-1/6} - y^{1/3})^9$ is given by:$T_{r+1} = ^9C_r (y^{-1/6})^{9-r} (-y^{1/3})^r$$T_{r+1} = ^9C_r (-1)^r y^{-(9-r)/6} y^{r/3}$$T_{r+1} = ^9C_r (-1)^r y^{(-9+r+2r)/6} = ^9C_r (-1)^r y^{(3r-9)/6}$For the term to be independent of $y$,the exponent of $y$ must be $0$:$(3r-9)/6 = 0$ $\Rightarrow 3r = 9$ $\Rightarrow r = 3$Substituting $r = 3$ into the general term expression:$T_{3+1} = ^9C_3 (-1)^3 y^0$$T_4 = - ^9C_3 = - \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = - 84$.

The term independent of $y$ in the expansion of $(y^{-1/6} - y^{1/3})^9$ is

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