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Question 131

The term independent of x in the binomial expansion of $$\left(\frac{3x^2}{2} - \frac{1}{3x}\right)^9$$ is

Solution

The power of x in the first term of expansion would be 18
The power of x in the second term would be 16(of the first term in original bracket) - 1 (of the second term in original bracket) = 15
The power of x in the third term would be 14 - 2 = 12
The power of x in the fourth term would be 12 - 3 = 9
The power of x in the fifth term would be 10 - 4 = 6
The power of x in the sixth term would be 8 - 5 = 3
The power of x in the seventh term would be 6-6 = 0

The seventh term would be independent of x

The value of seventh term would be (9C6)*$$\left(\frac{3}{2}\right)^3\cdot\left(\frac{1}{3}\right)^6$$ = 7/18

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