已知数列{an}满足a1=1,a2=2,a(n+2)=[an+a(n+1)]2,n属于正整数.令bn=a(n+1)-an,证{bn}是等比.求{an}通式

题目

已知数列{an}满足a1=1,a2=2,a(n+2)=[an+a(n+1)]2,n属于正整数.令bn=a(n+1)-an,证{bn}是等比.求{an}通式

答案

证明 令bn=a(n+1)-an
2a(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)
∴a(n+1)-an=(-1/2)^(n-1)
∴an-a(n-1)=(-1/2)^(n-2),
a(n-1)-a(n-1)=(-1/2)^(n-3)
……
a2-a1=(-1/2)^0
上面各式叠加得 an-a1=(-1/2)^0+……+(-1/2)^(n-3)+(-1/2)^(n-2)
=[1-(-1/2)^(n-1)]/(1+1/2)=(2/3)[1-(-1/2)^(n-1)]
∴an=a1+(2/3)[1-(-1/2)^(n-1)]=5/3-(2/3)*(-1/2)^(n-1)=5/3+(1/3)*(-1/2)^(n-2)