Question

已知数列{an},a1=5,(2n-1)an-8*n^2=(2n+1)an-1.求通项公式an

Answer 1

化简得 an/(2n+1)-a(n-1)/(2n-1)=8n^2/(4n^2-1)=2+[1/(2n-1)-1/(2n+1)]

设bn=an/(2n+1)

则b(n-1)=a(n-1)/(2n-1) b1=a1/3=5/3

bn-b(n-1)=2+[1/(2n-1)-1/(2n+1)]

b(n-1)-b(n-2)=2+[1/(2n-3)-1/(2n-1)]

.

b2-b1=2+(1-1/3)

叠加bn-b1=2n+[1-1/(2n+1)]

bn=2n-1/(2n+1)+8/3

an=(2n+1)bn=4n^2+22n/3+5/3