高中文科数学第五章 数 列

温馨提示此套题为Word请按住Ctrl,滑动鼠标滚轴调节合适观看比例，答解析附后。闭Word档返回原板块。课时分层作业三十等数及前和题每小题5分,共25分拟列a}的n为SnnS=8,a=()336C.由题意,等为q,由S=8,33得=2,q=2,所以=aq25=64,故选C.161国卷Ⅲ等差数{}的首项为1,公为0.若a,a,a列n236a}前6()nA.设等差数列的公差d,d≠d=-2d(d0),所所S=66

=aa,(26

2):“远层增一灯?”意思:座7了381灯且相的2倍则塔()

B.设塔的顶层共有x盏,则各层的灯数构成一个公比2的等,由=381可拟列}的前n项为=a·nn

n-1+,a的值()A.当n≥时,=S=a·nnn-1

n-1

-a·

n-2

=a·

n-2

,当n=1时a=S=a+,11a}是等比数列以a+=,所.n

列a}中若a)=,则cos(aa)是)n1425

D.B.由等比数列的通项公式可知aa2=2(aa),cos(a)=1-25141425

2

(aa)=1-2×=14题每小题5分,共15分考列a}和等比列b}足a=b=8,则nn1144】设等差数列{}的为d,等比数列{b}的公为q.由题意-1+3d=nn

3⇒

=:1列a}是列a=,则aaa+…+an25123234nn+1n+2列a}的q,q3==,解q===4.aaa}n1nn+1n+2aaa=4××1=8,公比为q=的列所aaaa+…12323234aa==nn+1n+2

-3n).

-3n

)设Sn

前n项和若a=1且3S,2S,S成等1123列an】3S,2S,S成等列q=3,然后求a123n】因为3S,2S成等列1232×2(a+a)=3a+a+a+a=3a12112332a=aqn1

n-1

n-1

.:3

n-1题每小题10分,共)拟列}中,a且,a列n1124求{}的通项公式及其前n项和.nn

设b=求数{}的前2n项和T.nn2n(1)设等差数列a}的为d,a=1,且a,a+2成等比.所n1124=a·(a+2),141+d)2=1×(1+3d+2),解得或d=-1,=0,舍2d=2,可anSn

2=n

.n为偶时=当为奇数时==.b}的奇数项是以项,n16为公列

列偶数8为首b}的前T=(b+b…+bn2n13

2n-1

)+(b+24+=2n

n

-n).编列{a}的n项和S,n∈nnn≥2时,+5S=8S+S.n+2nn-1a的值4

*.a=1,a=,a,123

证明

列(1)当n=24S+5S=8S+S,42314

+1,a=.4(2)+S(n≥n+2nn+1n-1+S(n≥2),n+2nn-1n+1n+an+2nn+14a=4×,312,n+2nn+1====,

以-a=1为公列21)(2018拟列a}足aa∈Nn3n3n+1a+a=9,则lo(a+a+a)的值是)246579B.-D.

*

A.因为loga+1=loga,所a=3a.3n3n+1n+1na}是公q=3列na+a+a=a(1+q22462

4a+a+a=a(1+q243(1+q2+q4)=3.5795253】等比数列a}满足a>0,n∈nn

*

,且a·a2n32n-3

≥2),则当n1a+loga…a212222n-1质得a=232n-3

2na=2nn

.所loga+loga+…2122=log)·a)·a)a]=log22n-1212n-122n-2n-1n+12

n(2n-1)

:2n2)已知数{}为等比数列若a+a=10,则a(a+2aa为()n4671339C.a(a+2a)+aa=aa+2aaa=a+)2=10713397173394646)(2016国卷)列a}足a+a=10,a+a则a…an132412n为】由于a,设a=an-1a是,是公比nn11

a,na·a·…·=12n=.n=3或4时,

值26

a·a·…·的12n:64分)(2016)列a}前n和S=1+λa中nnn明a}列,式n若S=,求λ.5(1)由题意得a=1+λ,故a1111

,S=1+λ,S=1+得a=λ-λ,=nnn+1n+1n+1n+1na}是以a=n1a=n(2)由(1)得Sn

,以.,又因为=,5

,即=,解λ=-1.分)(2018拟列a}足a=5,a=5,an12n+1nn-1证{a+2a列n+1n求{}的通项公式.n(1)因为a=a+6a(n≥2),n+1nn-1a+2a=3a+6a)(n≥2).n+1nnn-1nn-1a=5,a12a+2a=15,21a+2a0(nnn-1a+2a}是15,为公比的等比列n+1n(2)由(