Question

数列{An}中,A1=6,且An-A(n-1)=A(n-1)/n +n+1,求{An}通项公式?(n为整数且n>1)

Answer 1

An=n^2+3n+2

Answer 2

变形得
An=(n+1)/n *A(n-1)+n+1
两边同时除以n+1
An/(n+1)=A(n-1)/n +1
即数列{An/(n+1)}为公差为1的等差数列
从而 An/(n+1)=A1/2+(n-1)=n+2
从而 An=(n+1)(n+2)