设数列{an}的前n项和Sn=4/3an-(1/3)*(2^(n+1))+2/3, n=1,2,3……这题的第二问怎么做?设数列{an}的前n项和Sn=4/3an-(1/3)*(2^(n+1))+2/3, n=1,2,3……(1) 求首项a1与通项an(2)设Tn=(2^n)/Sn, n=1,2,3……,证明:T1+T2+T3+…Tn<
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![设数列{an}的前n项和Sn=4/3an-(1/3)*(2^(n+1))+2/3, n=1,2,3……这题的第二问怎么做?设数列{an}的前n项和Sn=4/3an-(1/3)*(2^(n+1))+2/3, n=1,2,3……(1) 求首项a1与通项an(2)设Tn=(2^n)/Sn, n=1,2,3……,证明:T1+T2+T3+…Tn<](/uploads/image/z/6997094-62-4.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%3D4%2F3an-%281%2F3%29%2A%282%5E%28n%2B1%29%29%2B2%2F3%2C+n%3D1%2C2%2C3%E2%80%A6%E2%80%A6%E8%BF%99%E9%A2%98%E7%9A%84%E7%AC%AC%E4%BA%8C%E9%97%AE%E6%80%8E%E4%B9%88%E5%81%9A%3F%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%3D4%2F3an-%281%2F3%29%2A%282%5E%28n%2B1%29%29%2B2%2F3%2C+n%3D1%2C2%2C3%E2%80%A6%E2%80%A6%281%29+%E6%B1%82%E9%A6%96%E9%A1%B9a1%E4%B8%8E%E9%80%9A%E9%A1%B9an%282%29%E8%AE%BETn%3D%282%5En%29%2FSn%2C+n%3D1%2C2%2C3%E2%80%A6%E2%80%A6%2C%E8%AF%81%E6%98%8E%EF%BC%9AT1%2BT2%2BT3%2B%E2%80%A6Tn%3C)
设数列{an}的前n项和Sn=4/3an-(1/3)*(2^(n+1))+2/3, n=1,2,3……这题的第二问怎么做?设数列{an}的前n项和Sn=4/3an-(1/3)*(2^(n+1))+2/3, n=1,2,3……(1) 求首项a1与通项an(2)设Tn=(2^n)/Sn, n=1,2,3……,证明:T1+T2+T3+…Tn<
设数列{an}的前n项和Sn=4/3an-(1/3)*(2^(n+1))+2/3, n=1,2,3……这题的第二问怎么做?
设数列{an}的前n项和Sn=4/3an-(1/3)*(2^(n+1))+2/3, n=1,2,3……
(1) 求首项a1与通项an
(2)设Tn=(2^n)/Sn, n=1,2,3……,证明:T1+T2+T3+…Tn< 3/2
第二问!?
设数列{an}的前n项和Sn=4/3an-(1/3)*(2^(n+1))+2/3, n=1,2,3……这题的第二问怎么做?设数列{an}的前n项和Sn=4/3an-(1/3)*(2^(n+1))+2/3, n=1,2,3……(1) 求首项a1与通项an(2)设Tn=(2^n)/Sn, n=1,2,3……,证明:T1+T2+T3+…Tn<
a1=2
Sn=4/3an-(1/3)*(2^(n+1))+2/3,
Sn-1=4/3a(n-1)-(1/3)*(2^n)+2/3,
相减得
an=4/3an-4/3a(n-1)-(1/3)*(2^n)
an=4a(n-1)+2^n
4an-1=4^2*a(n-2)+4*2^(n-1)
...
4^(n-2)a2=4^(n-1)*a1+4^(n-2)*2^2
以上叠加
an=4^(n-1)*a1+2^n+4*2^(n-1)+...+4^(n-2)*2^2
=2^(2n-1)+2^n*[2^(n-1)-1]
=2^(2n)-2^n
2)设Tn=(2^n)/Sn
Sn=4/3[2^(2n)-2^n]-(1/3)*(2^(n+1))+2/3
=4/3*2^(2n)-2^(n+1)+2/3
Tn=(2^n)/Sn
=1/[4/3*2^n-2+2/(3*2^n)]
=3/2*【1/(2^(n+1)+1/2^n-3)】
Tn<3/2n
放缩T1+T2+T3+…Tn