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Question
Answer the following:
Find third term in the expansion of `(9x^2 - y^3/6)^4`
Solution
Here a = 9x2, b = `(-y^3)/6`, n = 4
For 3rd term, r = 2
We have, tr+1 = nCr an–r .br
∴ t3 = `""^4"C"_2 (9x^2)^(4 - 2)((-y^3)/6)^2`
= `(4!)/(2!2!)(81x^4).y^6/36`
= `(4 xx 3)/(2 xx 1) xx 81x^4 xx y^6/36`
= `27/2 x^4y^6`
∴ 3rd term in the expansion of `(9x^2 - y^3/6)^4` is `27/2 x^4y^6`.
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