Question

数学（2x+1/x^2)^n的展开式中，第三项的二项式系数比第二项的二项式系数大27，求展开式中的常数项

Answer 1

C(2,n) - C(1,n) = 27
n(n-1)/(1*2) - n = 27
n(n-1) - 2n = 54
n(n-3) = 54 = 9×6
n = 9

(2x+1/x²)^n展开式中的常数项
= C(3,9) × (2x)^6 × (1/x²)³
= C(3,9) × 2^6 × x^6 × (1/x^6)
= C(3,9) × 2^6
= (9×8×7) / (1×2×3) × 2^6
= 3×4×7 × 2^6
= 3×7 × 2^8
= 21 × 256
= 5376

Answer 2

Cn 2-Cn 1=27
n(n-1)/2-n=27
n=9
常数项为(C9 3)2的6次方=5376