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प्रश्न
Answer the following:
Find the middle term (s) in the expansion of `((3x^2)/2 - 1/(3x))^9`
उत्तर
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,
tr+1 = nCr an-r br
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`.
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