数列{an}的前n项和记为Sn,a1=t,点(Sn,a(n+1))(n+1为底数)在直线y=2x+1上,n∈N+
(1.)当实数t为何值时,数列{an}是等比数列?
(2.)在(1.)的结论下,设bn=log3a(n+1),Tn是数列{1/bn·b(n+1)}的前n项和,求T2013的值
人气:160 ℃ 时间:2019-08-21 23:07:05
解答
1) (y1-y2)/(x1-x2)=2
a(n+1)-an = 2an
a(n+1)/an = 3
a2 = 3t =2a1+1 = 2t+1
t = 1
2)an = 3^(n-1)
a(n+1) = 3^n
bn = n
Tn = 1/2 +1/6+.
= 裂项法...能具体点吗?我觉得已经够详细了裂项法..1/1 * 1/2 = 1 -1/2.1/2 * 1/3 = 1/2 - 1/31/3 * 1/4 = 1/3 - 1/4...以此类推,当它们相加,1/2,1/3....全部抵消,只剩首尾即:T2013 = 1 - 1/2013 = 2012/2013
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