Find the term independent of `x` in the expansion of `(1+x+2x^3)[(3x^2//2)-(1//3)]^9`

Answer

`17/54`

Explanation

In the expansion of `E = ( 3/2x^(2) - 1/ (3x))^(9)` , we have
`T_(r+1) = (-1)^(r)*^(9)C_(r)*(3/2x^(2))(9-r) (1/(3x))^(r)`
`rArr T_(r+1) = (-1)^(r)*^(9)C_(r)*(3(9-2r))/(2^((9-r)))* x^((18-3r)).`
`(1+ x + 2x^(3)) [(a_(0) xx 1/x^(3)+a_(1) xx 1/x + a^(2))" from E"]`
`=(1+x+2x^(3)) [ {(-1)^(7)*^(9)C_(7)*3^(-5)/2^(2) xx1/(x^(3))}+ {(-1)^(6)*^(9)C_(6) * 3^(-3)/2^(3)xxx^(0)}]`
`[(x=-1 rArr18-3r = -1 rArr " r is faction"),(18-3r=0 rArr r=6 and 18-3r = -3 rArr r=7)]`
`=(1+ x+ 2x^(3))[(-1)/(27x^(3))+7/18]`
`:. " required term " ((-2)/27+7/18) = 17/54.`