Question

已知数列{an}中,a1=1,an=1/3a(n-1)-1/4,n>=2,求an

Answer 1

an=1/3a(n-1)-1/4
12an=4a(n-1)-3
12an+4.5=4a(n-1)+1.5
24an+9=8a(n-1)+3
3(8an+3)=8a(n-1)+3
3(8an+3)/[8a(n-1)+3]=1
(8an+3)/[8a(n-1)+3]=1/3
所以8an+3是以1/3为公比的等比数列

8an+3=(8*a1+3)*q^(n-1)
8an+3=(8+3)*(1/3)^(n-1)
8an+3=11*(1/3)^(n-1)
8an=11*(1/3)^(n-1)-3
an=[11*(1/3)^(n-1)-3]/8

Answer 2

an+3/8=1/3(a(n-1)+3/8)
an+3/8=(1/3)*(n-1)x(a1+3/8)=(1/3)*(n-1)x11/8
an=(1/3)*(n-1)x11/8-3/8