Question

数列{an}的前n项和为Sn=n的平方+2n+5 求an

Answer 1

Sn=n^2+2n+5

Sn-1=(n-1)^2+2(n-1)+5=n^2-2n+1+2n-2+5

=n^2+4

因为：

Sn=a1+a2+a3+…+an-1+an

Sn-1=a1+a2+a3+…+an-1

所以错位相减：an=Sn-Sn-1

则：an=Sn-Sn-1=n^2+2n+5-(n^2+4)

=n^2+2n+5-n^2-4

=2n+1