于是cn=lg[2000()]=3+lg2(n+)lg0.7數(shù)列{cn}是一個(gè)遞減的等差數(shù)列.因此.當(dāng)且僅當(dāng)cn≥0.且cn+1<0時(shí).數(shù)列{cn}的前n項(xiàng)的和最大. 查看更多

 

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已知數(shù)列{an}前n項(xiàng)和為Sn,且a1=2,3Sn=5an-an-1+3Sn-1(n≥2,n∈N*).
(Ⅰ)求數(shù)列{an} 的通項(xiàng)公式;
(Ⅱ)設(shè)bn=(2n-1)an,求數(shù)列{bn} 的前n項(xiàng)和為Tn;
(Ⅲ)若cn=tn[lg(2t)n+lgan+2](t>0),且數(shù)列{cn} 是單調(diào)遞增數(shù)列,求實(shí)數(shù)t的取值范圍.

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已知數(shù)列(an}為Sn且有a1=2,3Sn=5an-an-1+3Sn-1 (n≥2)
(I)求數(shù)列{an}的通項(xiàng)公式;
(Ⅱ)若bn=(2n-1)an,求數(shù)列{bn}前n和Tn
(Ⅲ)若cn=tn[lg(2t)n+lgan+2](0<t<1),且數(shù)列{cn}中的每一項(xiàng)總小于它后面的項(xiàng),求實(shí)數(shù)t取值范圍.

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已知數(shù)列{an} 的前n項(xiàng)和為Sn ,且有a1=2,3Sn=5an-an-1+3Sn-1(n≥2).
(1)若bn=(2n-1)an,求數(shù)列{bn}的前n項(xiàng)和Tn;
(2)若cn=tn[lg(2t)n+lgan+2](0<t<1),且數(shù)列{cn} 中的每一項(xiàng)總小于它后面的項(xiàng),求實(shí)數(shù)t的取值范圍.

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設(shè)正項(xiàng)數(shù)列{an}的前n項(xiàng)和為Sn,若{an}和{}都是等差數(shù)列,且公差相等.

(1)求{an}的通項(xiàng)公式;

(2)若a1,a2,a5恰為等比數(shù)列{bn}的前三項(xiàng),記數(shù)列cn,數(shù)列{cn}的前n項(xiàng)和為Tn.求證:對(duì)任意n∈N*,都有Tn<2.

 

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已知數(shù)列{an}滿足a1=0,a2=2,且對(duì)任意m、nN*都有
a2m1a2n1=2amn1+2(mn)2
(Ⅰ)求a3,a5
(Ⅱ)設(shè)bna2n1a2n1(nN*),證明:{bn}是等差數(shù)列;
(Ⅲ)設(shè)cn=(an+1an)qn1(q≠0,nN*),求數(shù)列{cn}的前n項(xiàng)和Sn.

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