=sin30°=.所以rn=rn-1(n≥2) 查看更多

 

題目列表(包括答案和解析)

(2013•房山區(qū)一模)已知全集U=R,集合M={x|x≤1},N={x|x2>4},則M∩(?RN)=( 。

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(2011•江西模擬)已知數(shù)列{an},{bn}分別是等差、等比數(shù)列,且a1=b1=1,a2=b2,a4=b3≠b4
①求數(shù)列{an},{bn}的通項(xiàng)公式;
②設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,求{
1
Sn
}的前n項(xiàng)和Tn;
③設(shè)Cn=
anbn
Sn+1
(n∈N),Rn=C1+C2+…+Cn,求Rn

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設(shè)函數(shù)f(x)=x2+1,g(x)=x,數(shù)列{an}滿足條件:對(duì)于n∈N*,an>0,且a1=1并有關(guān)系式:f(an+1)-f(an)=g(an+1),又設(shè)數(shù)列{bn}滿足bn=
log
a
an+1
(a>0且a≠1,n∈N*).
(1)求證數(shù)列{an+1}為等比數(shù)列,并求數(shù)列{an}的通項(xiàng)公式;
(2)試問(wèn)數(shù)列{
1
bn
}是否為等差數(shù)列,如果是,請(qǐng)寫出公差,如果不是,說(shuō)明理由;
(3)若a=2,記cn=
1
(an+1)-bn
,n∈N*,設(shè)數(shù)列{cn}的前n項(xiàng)和為Tn,數(shù)列{
1
bn
}的前n項(xiàng)和為Rn,若對(duì)任意的n∈N*,不等式λnTn+
2Rn
an+1
<2(λn+
3
an+1
)
恒成立,試求實(shí)數(shù)λ的取值范圍.

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(2011•自貢三模)己知.函數(shù)f(x)=
x-4
x+1
(x≠-1)的反函數(shù)是f-1(x).設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,對(duì)任意的正整數(shù)都有an=
f-1(Sn) -19
f-1(Sn)+1
成立,且bn=f-1(an)•
(I)求數(shù)列{bn}的通項(xiàng)公式;
(II)記cn=b2n-b2n-1(n∈N),設(shè)數(shù)列{cn}的前n項(xiàng)和為Tn,求證:對(duì)任意正整數(shù)n都有Tn
3
2
;
(III)設(shè)數(shù)列{bn}的前n項(xiàng)和為Rn,已知正實(shí)數(shù)λ滿足:對(duì)任意正整數(shù)n,Rn≤λn恒成立,求λ的最小值.

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設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,對(duì)任意的正整數(shù)n,都有an=5Sn+1成立,記bn=
4+an
1-an
(n∈N*)

(I)求數(shù)列{an}與數(shù)列{bn}的通項(xiàng)公式;
(Ⅱ)設(shè)數(shù)列{bn}的前n項(xiàng)和為Rn,是否存在正整數(shù)k,使得Rn≥4k成立?若存在,找出一個(gè)正整數(shù)k;若不存在,請(qǐng)說(shuō)明理由;
(Ⅲ)記cn=b2n-b2n-1(n∈N*),設(shè)數(shù)列{cn}的前n項(xiàng)和為Tn,求證:對(duì)任意正整數(shù)n都有Tn
3
2

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