【文档说明】广东省佛山市第一中学2021-2022学年高二下学期第一次段考试题（3月） 数学答案.docx，共(5)页，85.592 KB，由管理员店铺上传

转载请保留链接：https://www.doc5u.com/view-f3e33044e1ca1ab4d473b27009ecc470.html

以下为本文档部分文字说明：

佛山一中2020级高二下学期第一次段考数学答案及解析ABCACACACDADBBACDC.12.11.10.9.81−分）分第二个空（注：第一个空或3218;216.)31(32)2(cos41,915.114.48.132nnnnna−+++=−分

的通项公式为数列分解之得分依题设可得解-5-----------------1344)1(9}{-4---------------------------------------------4,92----------------

---------,1215511,4)1(:17.111115−=−+−==−==+=+=nnaadadaSdaann分分分分知由-10-------------------------------------------------------768012

9---------------------------------------------------21)21(2-7-------------------------222-6----------------------------------------

-----2)1()2(4449945921134−=−−=+++=+++==−−−−−nnnnnnnbbbTb.}{}{)3(}{:)2(;}{:)1(.434,434,0,1,}{}{.181111的通项公式和求数列是等差数列数列求证是等比数列数列求证中和已知数列nn

nnnnnnnnnnnnbababababbaababa−+−+−=+−===++分为公比的等比数列以为首项是以数列分分依题设可得解-4-----------21,1}{-3-------------------

------------------------)(212---------------------------)(2)(4)1(:111111=+++=++=+++++babababababann

nnnnnnnn分为公差的等差数列以为首项是以数列分分依题设可得-8-------------2,1}{-7------------------------------------------2)()(6---------------------

--------8)(4)(4)2(111111=−−=−−−+−=−++++babababababannnnnnnnnn分分分分得由-12---------------------------------------------).2121(2111-----

---------------------------------------),1221(2110---------------------------------------------------12-9---------------

-----------------------,21:)2(),1()3(111nbnanbabannnnnnnnn−+=−+=−=−=+−−−.}{),1(log)2(;}1{:)1(.23,2:}{.19311nnn

nnnnnnSnbaabaaaaa项和的前求数列记是等比数列数列求证满足已知数列+=++==+分分为公比的等比数列以为首项是以数列依题设可得解4-----------------------------

------------1331-3-----------3,31}1{)1(31)1(:11−==+=+++=++nnnnnnnaaaaaa分可得由-6-----------------)13(

)1(log)1()2(3−==+=nnnnnnbanab分分分分记-11-----------------------------------433)12(9----------------------------

------------331)31(3-8---------------------------333327-------------------33)1(32313332311112113221+−=−−−=−+++=−+−+++=+++=++++nnn

nnnnnnnnnnTnnTnnTnT分12-------------.2)1(433)12()21(1+−+−=+++−=+nnnnTSnnn;2022}{,1)2(;}11{:)1(.12,2:}{.20111项和的前求数列记是等差数

列数列求证满足已知数列+++=−−==nnnnnnnnbbnabaaaaa.35:,,12)3(212+++=−−=nnnnnScccSnac求证且记分分为公差的等差数列以为首项是以数列分依题设可得解-4----------------------------------------

-------------------11-3-----------------------------------------------1)1(1111,111}11{2-----------1111111112111)1(:11nannaaaaaaaaaannnnn

nnnnn+==−+=−=−−+−=−+−=−=−−=−+分分分可得由-8---------------2023202212022116---------------------------------111)1(1-5--------------------

---------------11)1()2(2023202232211=+−==++++−=+==+=+bbbbbbnnnnbbnnabnnnn分证毕时当分时当分可得由12----------

-------.3512235)12131(21)1211211321132112211221(21,210-----------------------------------------------351,1-9-

--------)121121(214444112)1()3(21112222+−=+−+=+−−+++−−++−−++++====+−−=−==−−=nnnncccSncSnnnnnnnacnnnn.}{,,,,21:}{,.211nnnnnnnnSnaamaaaaaNm

项和为的前数列记为奇数为偶数满足各项均为正整数的数列已知+=+.,25,5)3(;,7,32)2(;,2,8)1(132022171的值求若的值求若的值求若aSmSmaSma======分分分解：3---------------------------

----------------------------------30-2-----------------------------7,5,31---------------121,221,4212,8)1(76756453423121==+==+==

+=========Smaamaamaaaaaaaama分分-7-------------------------------.7623)1248(5051632-5-------------------------------------------------------,

121221,421,8,121,221421,821,16217,32)2(202291089786756453423121=+++++========+=============Saaaaaamaaaaaaaaaaaama分为正整数矛盾与时当-8-----------------

--------------------------------725,25721,221,)(4)3(323121kkkSkaakaaNkka========分时为正奇数当10-------------------105,25545,21,)(21323121===+=

+=+====akkSkmaakaakka分或综上分时当-12------------------------------------------------------107-11--------------------------------------,74,255

5221,42,)(1211323121====+=+==+=+=−=aakkSkaakmaaNkka.}22{,1)2(;)1(.9,5,3,}{,0}{.221121321nnnnnnnknTnbbnkbkkkkkkadan项和的前

求数列记求其中是等比数列且数列的等差数列不为是公差已知数列++−+−=+++===分分即有依题设可得解-2---------------------------------------0002-1-------)8)(2()4()1(:1111219325==++=+

=addadadadaaaa分为公比的等比数列以为首项是以数列5-----------------------------12222)1(2,2}{8,4,2121+===−====−nnnnnkkkkkkdddkadadadadannn分6------

-------------------------------------22121nkkknn+−=++++分分分可得由-12-------------------.12232]2)1(212121[29------------

-------------------]2)1(21221[2]2)1(2)[22(2222-7---------------------------221)1()2(11111111+−−=++−−+−=++−−+−=++−+−−=−+−=+−=++

+++++nnTnnnnbbnnkbnnnnnnnnnnnnnn获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com