Question

高二数学数列求通项公式问题～ 求详细过程

高二数学数列求通项公式问题～
求详细过程

Answer 1

a(n+1)= ( an+6 )/(3an -2)
a(n+1) -2 = ( an+6 )/(3an -2) -2
= (-5an+10)/(3an-2)
1/[a(n+1) -2] = -(1/5) . [(3an -2) /(an -2) ]
=-(1/5)[ 3 + 4/(an -2) ]
= -(4/5)[ 1/(an-2) ] - 3/5
1/[a(n+1) -2] + 1/3 = -(4/5) [ 1/(an-2) + 1/3]

=> {1/[an -2] + 1/3} 是等比数列, q= -4/5
1/(an -2) + 1/3 = (-4/5)^(n-1) . (1/(a1 -2) + 1/3)
=-(2/3). (-4/5)^(n-1)
an =1/[ -1/3 -(2/3). (-4/5)^(n-1) ] + 2

Answer 2

听说过特征方程吗？

追问

没..