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  • 數(shù)列{an}的前n項(xiàng)和記為Sn,a1=t,點(diǎn)(Sn,a(n+1))(n+1為底數(shù))在直線y=2x+1上,n∈N+

    數(shù)列{an}的前n項(xiàng)和記為Sn,a1=t,點(diǎn)(Sn,a(n+1))(n+1為底數(shù))在直線y=2x+1上,n∈N+
    (1.)當(dāng)實(shí)數(shù)t為何值時(shí),數(shù)列{an}是等比數(shù)列?
    (2.)在(1.)的結(jié)論下,設(shè)bn=log3a(n+1),Tn是數(shù)列{1/bn·b(n+1)}的前n項(xiàng)和,求T2013的值
    數(shù)學(xué)人氣:290 ℃時(shí)間:2019-08-21 23:07:05
    優(yōu)質(zhì)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+.
    = 裂項(xiàng)法...能具體點(diǎn)嗎?我覺得已經(jīng)夠詳細(xì)了裂項(xiàng)法..1/1 * 1/2 = 1 -1/2.1/2 * 1/3 = 1/2 - 1/31/3 * 1/4 = 1/3 - 1/4...以此類推,當(dāng)它們相加,1/2,1/3....全部抵消,只剩首尾即:T2013 = 1 - 1/2013 = 2012/2013
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