Question

Answer the following:

Find the middle term (s) in the expansion of `((3x^2)/2 - 1/(3x))^9`

Answer 1

There are 10 terms in the expansion of `((3x^2)/2 - 1/(3x))^9`.

∴ middle terms are t₅ and t₆.

We know that, in the expansion of (a+ b)ⁿ,

t_r+1 = ⁿC_r a^n-r b^r 

Here a = `(3x^2)/2`, b = `-1/(3x)`, n = 9

For t₅, r = 4 and for t₆, r = 5

∴ t₅ = `""^9"C"_4 ((3x^2)/2)^(9 - 4) (-1/(3x))^4` 

= `(9 xx 8 xx 7 xx 6)/(1 xx 2 xx 3 xx 4) xx (3^5.x^10)/32 xx 1/(3^4.x^4)`

= `(189x^6)/16`

and t₆ = `""^9"C"_5 ((3x^2)/2)^(9 - 5) (-1/(3x))^5`

= `(9 xx 8 xx 7 xx 6 xx 5)/(1 xx 2 xx 3 xx 4 xx 5) xx (3^4.x^8)/16 xx (-1)/(3^5.x^5)`

= `(-21x^3)/8`

∴ the middle terms are `(189x^6)/16` and `(-21x^3)/8`.