Question

数列{an}中,a1=2,an+1=an+4n,求通项公式an

Answer 1

(1)an+1=an+4n

an=an-1+4(n-1)

an-1=an-2+4(n-2)

……＝……

a3=a2+4*2

a2=a1+4*1

等式左右分别相加

an+1+an+……+a3+a2=an+an-1+……+a2+a1+4n+4(n-1)+……+4*2+4*1

左右的an+……+a3+a2可抵消得

an+1=a1+4n+4(n-1)+……+4*2+4*1

=a1+4(n+n-1+……+2+1)

所以 an =2+4*（n-1+……+2+1)

＝2+4*（n-1+1)*(n-1)/2

=2+2n(n-1)

=2n^2-2n+2