已知数列{an}中,a1=1,a2=5/3,a(n+2)=5/3a(n+1)-2/3an,求数列{an}
人气：160 ℃ 时间：2020-02-15 08:48:55
解答
由 a(n+2) = 5/3a(n+1) - 2/3an可得,a(n+2) - a(n+1) = 2/3[a(n+1)-an][a(n+2)-a(n+1)] / [a(n+1)-an] = 2/3设 [a(n+1)-an] = bn则 b(n+1)/bn = 2/3b1 = a2 - a1 = 2/3所以,bn = 2/3 *(2/3)^(n-1) = (2/3)^n所以,a(n...
推荐
- 已知数列{an}满足a1=1/3.a2=7/9.a[n+2]=4/3a[n+1]-1/3an
- 设数列〔an〕满足a1=1,a2=5/3(3分之5）,an+2=5/3an+1－2/3an,(n属于N※）.
- 在数列{an}中,已知a1=4/3,a2=13/9,当n≥2,且n∈N*时,有a（n+1)=4/3an-1/3a(n-1)
- 已知数列{an}中,a1=1,a2=5/3,a(n+2)=5/3a(n+1)-2/3an,求数列{an}的前n项和Sn
- 在数列{an}中,a1=1,a2=2,a(n+2)=2/3a(n+1)+1/3an,求an+2)=2/3a(n+1)+1/3an,求an
- 量杯怎么能量面粉
- 已知函数f(x)=x^2+px+q,集合A={x/f(x)=x)},集合B={x/f[f(x)]=x}
- 圆的面积和半径_比例．
猜你喜欢
