Question

Find the middle term (terms) in the expansion of `(3x - x^3/6)^9`

Answer 1

Given expression is `(3x - x^3/6)^9`

Number of terms = 9 + 1 = 10  ....(even)

∴ Middle terms are `n^"th"/2` term and `(n/2 + 1)^"th"` term

= `10^"th"/2`

= 5^th term and (5 + 1) = 6^th term

General Term `"T"_(r + 1) = ""^n"C"_r  x^(n - r) y^r`

∴ T₅ = `"T"_(4 + 1)`

= `""^9"C"_4  (3x)^(9 - 4)  (- x^3/6)^4`

= `""^9"C"_4  (3)^5 * x^5  (-1/6)^4 * x^12`

= `(9 xx 8 xx 7 xx 6)/(4 xx 3 xx 2 xx 1) xx (3 xx 3 xx 3 xx 3 xx 3)/(6 xx 6 xx 6 xx 6) x^17`

= `189/8 x^17`

Now, T₆ = T₅₊₁

= `""^9"C"_5  (3x)^(9 - 5) (- x^3/6)^5`

= `""^9"C"_5  (3)^4 x^4 (- 1/6)^5 * x^15`

= `(9 xx 8 xx 7 xx 6 xx 5)/(5 xx 4 xx 3 xx 2 xx 1) (3)^4 (- 1/6)^5 * x^19`

= ` - 21/16 x^19`

Hence, the required middle terms are `189/8 x^17` and `- 21/16 x^19`