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

∵[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

d=a2-a1=3
a(n+2)=2a(n+1)-an+2
a(n+2)-a(n+1)=a(n+1)-an+2=3
an=a(n+1)-1

a（n+2)=a(n+1)+a(n+1)-an+2
a（n+2)-a(n+1)=a(n+1)-an+2
[a（n+2)-a(n+1)]-[a(n+1)-an]=2 即数列{a(n+1)-an}成等差数列
设bn=a(n+1)-an
b1=4-1=3
bn=3+2(n-1)=2n+1
a(n+1)-an=2n+1
an=a1+(a2-a1)+(a3-a2)+.....(an-a(n+1))
=1+3+5+.....2n+1
=(1+2n+1)(n+1)/2
=(n+1)(n+1)