Question

高中数列问题

a1=1. (3n-1)a(n+1)=(3n+2)an +1 ，求an

Answer 1

已知 a1=1
令n=1 a2=3
n=2 a3=5
n=3 a4=7
……
以此类推：an=2(n+1)

Answer 2

(3n-1)a(n+1)=[3(n+1)-1]an+1
a(n+1)/[3(n+1)-1]=an/(3n-1)+1/[3(n+1)-1](3n-1)

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

[a(n+1)+1/3]/[3(n+1)-1]=[(an+1/3)]/[(3n-1)]
数列{[(an+1/3)]/[1/(3n-1)]}各项均相等。
(an+1/3)/(3n-1)=(a1+1/3)/(3*1-1)=(1+1/3)/2=2/3

3(an+1/3)=2(3n-1)
3an+1=6n-2
3an=6n-3

an=2n-1

Answer 3

(3n-1)a(n+1)=(3n+2)an +1
故 (3n-1)[a(n+1)+1/3]=(3n+2)[an+1/3]
构造新数列{bn}，令bn=an+1/3，则
(3n-1)b(n+1)=(3n+2)bn
故b(n+1)=[(3n+2)/(3n-1)]bn
={[3(n+1)-1]/(3n-1)}bn
因此，bn={[3n-1]/[3(n-1)-1]}b(n-1)
={[3n-1]/[3(n-1)-1]}{[3(n-1)-1]/[3(n-2)-1]}b(n-2)
={[3n-1]/[3(n-2)-1]}b(n-2)
=...
={[3n-1]/[3*1-1]}b1
=[(3n-1)/2](a1+1/3)
=[(3n-1)/2](1+1/3)
=[2(3n-1)]/3
an=bn-1/3=[2(3n-1)]/3-1/3