MixedMediumQuestion 280014
Question
The term containing x¹⁰ in (x² + 1/x)¹⁴ is:
- AT₅ = ¹⁴C₄ x¹⁰Correct
- BT₄
- CT₆
- DNo such term
Correct answer
T₅ = ¹⁴C₄ x¹⁰
Explanation
T_{r+1} = ¹⁴Cᵣ (x²)^(14−r) (1/x)^r = ¹⁴Cᵣ x^(28−2r−r) = ¹⁴Cᵣ x^(28−3r). For x¹⁰: 28−3r = 10 → r = 6. T₇ = ¹⁴C₆ x¹⁰ = 3003x¹⁰. So T₇, not T₅.