The absolute difference of the coefficients o | vedclass

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Question

$T _{ r +1}={ }^{11} C _{ r }\left(2 x ^2\right)^{11- r }\left(\frac{1}{2 x }\right)^{ T }$

$={ }^{11} C _{ r } 2^{11-2 r } x ^{22-3 r }$

$22-3

Vedclass · Accepted Answer

$12^3-12$

Vedclass · Answer

$T _{ r +1}={ }^{11} C _{ r }\left(2 x ^2ight)^{11- r }\left(\frac{1}{2 x }ight)^{ T }$

$={ }^{11} C _{ r } 2^{11-2 r } x ^{22-3 r }$

$22-3 r =10 \quad 	ext { and } \quad 22-3 r =7$

$r =4 \quad 	ext { and } \quad r =5$

$	ext { Coefficient of } x ^{10}={ }^{11} C _4 \cdot 2^3$

$	ext { Coefficient of } x ^7={ }^{11} C _5 \cdot 2^1$

$	ext { difference }={ }^{11} C _4 \cdot 2^3-{ }^{11} C _5 \cdot 2$

$=\frac{11 	imes 10 	imes 9 	imes 8}{24} 	imes 8-\frac{11 	imes 10 	imes 9 	imes 8 	imes 7}{120} 	imes 2$

$=11 	imes 10 	imes 3 	imes 8-11 	imes 3 	imes 4 	imes 7$

$=11 	imes 3 	imes 4 	imes(20-7)$

$=11 	imes 12 	imes 13$

$=12(12-1)(12+1)$

$=12\left(12^2-1ight)$

$=12^3-12$

The absolute difference of the coefficients of $x^{10}$ and $x^7$ in the expansion of $\left(2 x^2+\frac{1}{2 x}\right)^{11}$ is equal to

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