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Binomial Theorem · JEE Preparation Mathematics

What is the term independent of x in the expansion of (x^2 + 1/x)^9?

Practice question from Binomial Theorem (JEE Preparation Mathematics), with the correct answer and a step-by-step solution.

medium · MCQ · Q-E7B04E

What is the term independent of x in the expansion of (x^2 + 1/x)^9?

  1. A.36
  2. B.84Correct answer
  3. C.126
  4. D.252

Solution

The general term is Tr+1=(9r)(x2)9−r(x−1)r=(9r)x18−3rT_{r+1}={9 \choose r}(x^2)^{9-r}(x^{-1})^r={9 \choose r}x^{18-3r}. For a term independent of xx, 18−3r=018-3r=0, so r=6r=6. Hence, the required term is (96)=84{9 \choose 6}=84.

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