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JEE Main 2023 · Mathematics · Binomial Theorem · medium

In the expansion of (2x^2-1/x)^6, the constant term is

Single correct option

In the expansion of (2x2−1x)6\left(2x^2-\dfrac1x\right)^6, the constant term is
  1. A. −60-60
  2. B. 6060
  3. C. 1515
  4. D. 240240
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Answer: (B) 6060

Tr+1=(6r)(2x2)6−r(−1x)r=(6r)26−r(−1)rx12−3rT_{r+1}=\binom6r (2x^2)^{6-r}\left(-\tfrac1x\right)^r=\binom6r2^{6-r}(-1)^r x^{12-3r}. The constant term needs 12−3r=012-3r=0, so r=4r=4. The term is (64)⋅22⋅(−1)4=15⋅4=60\binom64\cdot2^2\cdot(-1)^4=15\cdot4=60.
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