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

Explanation

We have,
`(1+x+2x^(3)) (3/2 x^(2) - 1/(3x))^(9)`
` = (1+x+2x^(3))[(3/2x^(2))^(9) - .^(9)C_(1)(3/2x^(2)) (1/(3x))+"……." + (-1)^(9)(1/(3x))^(9)] " "(1)`
Therefore, the term independent of x in the expansion is
`1a_(0) + 1a_(1) + 2a_(3) " "(2)`
where `a_(m)` is the coefficient of `x^(m)` in the second bracket `[]` of `(1)`.
Now, `(r+1)` th term in `[]` of `(1)` is
`.^(9)C_(r) (3/2x^(2))^(9-r)(-1//3x)^(r)=(-1)^(r).^(9)C_(r)(3/2)^(9-r)(1/(3'))(x^(18-3r))`
`:. a_(18x-3r) = "Coefficient of" x^(18-3r)`
` = (-1)^(r).^(9)C_(r)(3/2)^(p-r)(1//3^(r))`
Now, for `a_(0)`,
`18-3r=0` or `r = 6`
`rArr a_(0) = (-1)^(6).^(9)C_(6)(3/2)^(9-6) xx (1)/(3^(6)) = 7/18`
For `a_(1)`,
`18-3r = - 1`
or `r = 19//3`, which is fractional
`:. a_(1) = 0`
For `a_(3)`,
`18-3r = - 3` or `r = 7`
`rArr a_(3) = (-1)^(7) .^(9)C_(7)(3/2)^(9-7) 1/(3^(7)) = - (1)/(27)`
Hence from (2), the required term is
`1 xx 7/18 + 0 + 2 xx (- (1)/(27)) = (17)/(52)`.