数列an中,a1=1,sn满足sn+1=sn-1/2an,求an

数列an中,a1=1,sn满足sn+1=sn-1/2an,求an

详细过程，谢谢

s(n+1)=sn-1/2an,

s(n+2)=s(n+1)-1/2a(n+1)

a(n+2)=a(n+1)-1/2a(n+1)+1/2an

a(n+2)=1/2a(n+1)+1/2an

a(n+2)-a(h+1)=-1/2a(n+1)+1/2an=-1/2[a(n+1)-an]

{a(n+1)-an}为等比数列

a1=1

s2=s1-1/2a1=1/2a1=1/2

a2=-1/2

a2-a1= -3/2

a(n+1)-an=(-1/2)^(n-1)*(a2-a1)=-3/2*(-1/2)^(n-1)=3*(-1/2)^n

an-a(n-1)=3*(-1/2)^(n-1)

a(n-1)-a(n-2)=3*(-1/2)^(n-2)

....

a2-a1=3*(-1/2)^1

a(n+1)-a1=[3*(-1/2)^n+3*(-1/2)^(n-1)+.;...\]

a(n+1)-a1=(-3/2)*(1-(-1/2)^n)/(1-(-1/2))

=(-1/2)^n-1

a(n+1)-1=(-1/2)^n-1

a(n+1)=(-1/2)^n

an=(-1/2)^(n-1)

因为  s(n+1)-sn=a(n+1)

所以sn+1=sn-1/2an   得    a(n+1)=-1/2an

a(n+1)/an=-1/2

所以数列an=(-1/2)^(n-1)

谢谢采纳