数列{an}中,a1=-13,a（n+1）=（2an +3）/an,求{an}的通项公式.注:等号左

数列{an}中,a1=-13,a（n+1）=（2an +3）/an,求{an}的通项公式.注:等号左

a(n+1)=(2an+3)/ana(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+4an=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].======以下答案可供参考======供参考答案1:a(n+1)=(2an+3)/an1+a(n+1)=1+(2an+3)/an =(3an+3)/an所以1/[1+a(n+1)]=an/(3+3an) =1/3-1/3*[1/(1+an)]那么1/[1+a(n+1)]-1/4=-1/3*[1/(1+an)-1/4]而1/(1+a1)-1/4=1/(1-13)-1/4=-1/12-1/4=-1/3≠0所以数列{1/(1+an)-1/4}是以-1/3为首项、-1/3为公比的等比数列1/(1+an)-1/4=(-1/3)^n，所以an=-[(-3)^(n+1)+4]/[(-3)^n+4] (n∈N+)供参考答案2:a(n+1)=(2an+3)/ana(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+4an=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]。

这下我知道了