Question

已知数列an满足a1=1，a2=2，a(n+2)=an+a(n+1)/2

已知数列an满足a1=1，a2=2，a(n+2)=an+a(n+1)/2(1)令bn=a(n+1)-an，证明bn是等比数 (2)求an的通项公式

Answer 1

(1)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；(2)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；累加得：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)