The term independent of x in (x² + 1/x)⁶ is:
Solution
Term T(r+1) = C(6,r) (x²)^(6-r) (1/x)^r = C(6,r) x^(12-2r-r) = C(6,r) x^(12-3r). For x⁰: 12-3r=0 → r=4. Wait: r=4 gives C(6,4) = 15. Actually 12-3r=0 → r=4. C(6,4)=C(6,2)=15. The x-independent term is C(6,4) × (1)⁴ × (1)²? No: (x²)^(2) × (1/x)^(4) = x⁴/x⁴ = 1. So coefficient = C(6,4) = 15 = C(6,2) = C(6,3)? C(6,3)=20. So answer = 15, which is C(6,3) is wrong. C(6,4)=C(6,2)=15. The answer option a says C(6,3)=20 which is wrong. Option c says C(6,2)=15. So answer is c.
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: General Term
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