Question

一道小小的数列题数列{an}中.a1=1.a2=4.a(n+2)=2a(n+1)-an+2.求an.PS:a(n+2)表

Answer 1

∵[a(n+2)-a(n+1)]-[a(n+1)-an]=2

令bn=a(n+1)-an

∴bn为等差数列 又b1=a2-a1=3

∴bn=3+2(n-1)=2n+1

∴a(n+1)=an+2n+1

an=a(n-1)+2(n-1)+1

······

a2=a1+2(2-1)+1

∴a(n+1)=(1+n)n+n+1=（n+1）²

∴an=n² n≥1