已知数列(an)满足a1=1,a2=2,a(n+2)=1/2(an+a(n+1)),n属于自然数

令bn=a(n+1)-an2a(n+2)=an+a(n+1)∴2[a(n+2)-a(n+1)]=an-a(n+1)=-[a(n+1)-an]bn=a(n+1)-an,∴2b(n+1)=-bn,即b(n+1)/bn=-1/2∴{bn}是等比数列b1=a2-a1=2-1=1,{bn}是首项为1,公比为-1/2的等比数列∴bn=1*(-1/2)^(n-1)...