Coefficient of x^{11} in the expansion of lef | vedclass

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Question

Coefficent of $x^{11} \equiv \frac{\left(1+x^2\right)^4\left(1+x^3\right)^7\left(1+x^4\right)^{12}\left(1-x^2\right)^4}{\left(1-x^2\right)^4}$

Coef

Vedclass · Accepted Answer

$1113$

Vedclass · Answer

Coefficent of $x^{11} \equiv \frac{\left(1+x^2ight)^4\left(1+x^3ight)^7\left(1+x^4ight)^{12}\left(1-x^2ight)^4}{\left(1-x^2ight)^4}$

Coefficent of $x^{11} \equiv\left(1-x^8ight)^4\left(1+x^4ight)^8\left(1+x^3ight)^7\left(1-x^2ight)^{-4}$

$=\left(1-4 x^8ight)\left(1+x^4ight)^8\left(7 x^3+35 x^9ight)\left(1-x^2ight)^{-4} $

$=\left(7 x^3+35 x^9-28 x^{11}ight)\left(1+x^4ight)^8\left(1-x^2ight)^{-4}$

Coefficent of $x^8=\left(7 x+35 x^6-28 x^8ight)\left(1+8 x^4+28 x^8ight)\left(1-x^2ight)^{-4}$

$=\left(7+35 x^6-28 x^3+56 x^4+196 x^8ight)\left(1-x^2ight)^{-4}$

Coefficent of $t ^4 \equiv\left(7+56 t ^2+35 t ^3+168 t ^4ight)(1- t )^{-4}$

$=7 \cdot{ }^7 C _3+56 \cdot{ }^5 C _3+35 \cdot{ }^4 C _3+168 $

$=245+700+168=1113 .$

$Alterantive :$

$2 x+3 y+4 z=11 $

$(x, y, z)=(0,1,2){ }^4 C_0 	imes{ }^7 C_1 	imes{ }^{12} C_2 $

$(1,3,0)^4 C_1 	imes{ }^7 C_3 $

$(2,1,1)^4 C_2 	imes{ }^7 C_1 	imes{ }^{12} C_1 $

$(4,1,0)^7 C_1 $

$	ext { coefficient of } x^{11} =66 	imes 7+35 	imes 4+42 	imes 12+7 $

$\quad=1113$ Ans.

Coefficient of $x^{11}$ in the expansion of $\left(1+x^2\right)^4\left(1+x^3\right)^7\left(1+x^4\right)^{12}$ is

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