Question

已知数列{an}满足，a1=1，a2=2, a(n+2)=[an+a(n+1)]/2,令bn=a（n+1)-an，证明：{bn}是等比数列

（2）求{an}的通项

Answer 1

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)