Question

已知数列an满足a1=5/6,a(n+1)=1/3an+(1/2)^(n+1),n属于N*,数列bn满足bn=a（n+1

Answer 1

(1)bn=a(n+1)-1/2an,b(n+1)=a(n+2)-1/2a(n+1)

so b(n+1)/bn=……

将b(n+1)和bn中的a(n+2) a(n+1)和an全部化为an,可得b(n+1)/bn=1/3,得证

(2)a2=1/3a1+1/4=19/36,b1=a2-1/2a1=1/9,

由（1）知,bn为等比数列,公比为1/3,所以bn的前n项和Sn=b1*(1-(1/3)^n)/(1-1/3)=(1/6)*(1-(1/3)^n)

因为bn=a(n+1)-1/2an=-1/6an+(1/2)^(n+1)=b1*(1/3)^(n-1)=(1/3)^(n+1),

所以an=6[(1/2)^(n+1)-(1/3)^(n+1)]