设数列（An)满足A0=0A1=2.且对一切n属于N.有An+2=2An+1－An＋2.求An的通项公式

Question

设数列（An)满足A0=0 A1=2.且对一切n属于N.有An+2=2An+1－An＋2.求An的通项公式

Answer 1

a(n+2)=2a(n+1)-a(n) + 2
a(n+2)-a(n+1) = a(n+1) -a(n) +2
[a(n+2)-a(n+1)] -[a(n+1) -a(n)] =2
[a(n+1) -a(n)] - [a1-a0] = 2n
a(n+1) -a(n) = 2n+2
a(n)-a(n-1) = 2n
an - a1 = 4+6+8+..+2n
= (n+2)(n-1)
an = n^2+n