由an+1=2nan,得 an+1 an =2n,∴n≥2时, an an?1 =2n-1,∴n≥2时,an=a1× a2 a1 × a3 a2 ×…× an an?1 =1×2×22×…×2n-1=21+2+…+(n-1)=2 n(n?1) 2 ,又a1=1适合上式,∴an=2 n(n?1) 2 .