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Question
Find the constant term (term independent of x) in the expansion of `(2x + 1/(3x^2))^9`
Solution
Let tr+1 be the constant term in the expansion of `(2x + 1/(3x^2))^9`.
We know that, in the expansion of (a + b)n,
tr+1 = nCr an–r br
Here a = 2x, b = `1/(3x^2)`, n = 9
∴ tr+1 = `""^9"C"_"r" (2x)^(9-"r") (1/(3x^2))^"r"`
= `""^9"C"_"r" 2^(9 - "r")*x^(9 - "r")*(1/3)^"r"*x^(-2"r")`
= `""^9"C"_"r" 2^(9 - "r")*(1/3)^"r"*x^(9 - 3"r")`
But tr+1 is a constant term
∴ power of x = 0
∴ 9 – 3r = 0
∴ r = 3
∴ the constant term = `""^9"C"_3*2^(9 - 3)*(1/3)^3`
= `(9 xx 8 xx 7)/(1 xx 2 xx 3) xx 2^6 xx 1/3^3`
= `84 xx 64 xx 1/27`
= `1792/9`
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