Question

已知数列{a0}满足a0=1,an=a0+a1+..+an-1(n≥1),则n≥1时,an等于

Answer 1

由已知得,

an=a0+a1+...+an-1,(1)所以

an+1=a0+a1+...+an-1+an.(2)

(2)-(1)得an+1-an=an,n≥1

所以,an+1=2an,an+1/an=2,a1=2a0=2即

数列是以2为首项2为公比的等比数列,

an=a1q^(n-1)=2^n,n≥1.