数列{an}中,a1=-13,a（n+1）=（2an +3）/an,求{an}的通项公式.注:等号左边的括号表示角标

题目

数列{an}中,a1=-13,a（n+1）=（2an +3）/an,求{an}的通项公式.注:等号左边的括号表示角标

答案

a(n+1)=(2an+3)/an
a(n+1)+1=(3an+3)/an=3(an +1)/an (1)
a(n+1) -3=(2an+3-3an)/an=(-an +3)/an=-(an -3)/an (2)
(1)/(2)
[a(n+1)+1]/[a(n+1)-3]=(-3)(an +1)/(an -3)
{[a(n+1)+1]/a(n+1)-3}/[(an +1)/(an-3)]=-3,为定值.
(a1+1)/(a1-3)=(-13+1)/(-13-3)=3/4
数列{(an +1)/(an -3)}是以3/4为首项,-3为公比的等比数列.
(an +1)/(an -3)=(3/4)×(-3)^(n-1)=-(-3)ⁿ/4
-(-3)ⁿ×(an-3)=4an+4
an=3 -16/[(-3)ⁿ+4]
n=1时,an=3- 16/(-3+4)=3-16=-13同样满足.
数列{an}的通项公式为an=3 -16/[(-3)ⁿ+4].