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प्रश्न
Find the coefficients of x6 in the expansion of `(3x^2 - 1/(3x))^9`.
उत्तर
Let tr+1 contains x6 in the expansion of `(3x^2 - 1/(3x))^9`
We know that, in the expansion of (a+ b)n,
tr+1 = nCr an–r br
Here a = 3x2, b = `-1/(3x)`, n = 9
∴ tr+1 = `""^9"C"_"r"(3x^2)^(9-"r")(-1/(3x))^"r"`
= `""^9"C"_"r".3^(9 - "r").x^(18 - 2"r").(-1/3)^"r".x^(-"r")`
= `""^9"C"_"r".3^(9 - "r").(-1/3)^"r".x^(18 - 3"r")`
But tr+1 has x6
∴ power of x = 6
∴ 18 – 3r = 6
∴ r = 4
∴ coefficient of x6 = `""^9"C"_4.(3)^5(-1/3)^4`
= `(9 xx 8 xx 7 xx 6)/(1 xx 2 xx 3 xx 4) xx 3`
= 378
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