高一数学关于递推数列,帮帮忙拉~~~急1.在数列｛An｝中,a1=1在下列条件中,分别求通项公式;(1)3a(n+1)^2=2an^2+1(2)a(n+1)=an+1/√n+√(n+1)(3)na(n+1)=(n+1)an(n+1)(4)an+1=-2an-3(5)3an^2+2ana(n+1)-an^2(6)Sn=2an-2

Answer 1

(1)3a(n+1)^2-3=2an^2-2a(n+1)^2-1=2/3(an^2-1)a1=1?an=1如果a1=k不等于1令bn=an^2-1则b1=k^2-1b(n+1)=2/3bnbn=b1(2/3)^(n-1)=(k^2-1)*(2/3)^(n-1)an=√[(k^2-1)*(2/3)^(n-1)+1](2)1/√n+√(n+1)= √(n+1)-√na(n+1)...