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Question
Find the middle term (terms) in the expansion of `(3x - x^3/6)^9`
Solution
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`
= 5th term and (5 + 1) = 6th 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`
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