Question

a1=20,an+1=an+2n-1.求an的通项公式

Answer 1

an+1=an+2n-1

an+1-an=2n-1

an-a(n-1)=2(n-1)-1

a(n-1)-a(n-2)=2(n-2)-1

a(n-2)-a(n-3)=2(n-3)-1

.

a3-a2=2(2-1)-1=1

a2-a1=2(1-1)-1=-1

以上等式相加得

an-a1=2(n-1)-1+2(n-2)-1+2(n-3)-1+.+1-1

=[-2+2(n-1)]*(n-1)/2-n*1

=2(n-2)*(n-1)/2-n

=(n-2)*(n-1)-n

=n^2-3n+2-n

=n^2-4n+2

an=n^2-4n+2+a1

=n^2-4n+2+20

=n^2-4n+22