若(x+1x−2)n的展开式中常数项为-20，则自然数n=．

Question

若(x+(1)/(x)−2)n的展开式中常数项为-20，则自然数n=______．

Answer 1

(x+(1)/(x)−2)n=(√(x)−(1)/(√(x))) 2n
(√(x)−(1)/(√(x)))2n展开式的通项为Tr+1＝

C

r 2n

(√(x))2n−r(−(1)/(√(x)))r=（-1）^rC_2n^rx^n-r
令n-r=0得n=r
∴展开式中常数项为（-1）ⁿC_2nⁿ
∵展开式中常数项为-20
∴（-1）ⁿC_2nⁿ=-20
∴n=3
故答案为3