2013屆數(shù)列復(fù)習(xí)題（含答案）

1、1數(shù)列復(fù)數(shù)列復(fù)習(xí)題習(xí)題1.已知等差數(shù)列的首項(xiàng)及公差均為正數(shù)令當(dāng)是數(shù)列na2012(2012).nnnbaanNn?????kb的最大項(xiàng)時(shí)求的值nbk2.已知數(shù)列滿足求的前項(xiàng)和na1(1)21nnnaan?????na60解可證明:14142434443424241616nnnnnnnnnnbaaaaaaaab???????????????????112341515141010151618302baaaaS????????????3.已知

2、是等差數(shù)列其前項(xiàng)和為，是等比數(shù)列且=.nannSnb1a1=2b44=27ab44=10Sb?(Ⅰ)求數(shù)列與的通項(xiàng)公式nanb(Ⅱ)記證明.1121=nnnnTababab??nN?12=210nnnTab?()nN?解(1)設(shè)等差數(shù)列的公差為等比數(shù)列的公比為由得??nad??nbq112ab??由條件得方程組故344423286adbqSd?????33232273286210dqdqdq???????????????????312(

3、)nnnanbnN????（2）1211223112112222()22nnnnnnnnnnnaaTababababaaaa????????????????????111213132352222nnnnnnnannncc?????????????12231112[()()()]2()nnnnnnTcccccccc????????????1022(35)1021212102nnnnnnnbaTba???????????方法二：數(shù)學(xué)歸納法（

4、1）當(dāng)時(shí)，，故等式成立。1n?1111112121621016Tabab???????3211[2(1)(2)1kkkaqqqq????????21(2)01kaqqqq?????因此對(duì)任意成等差數(shù)列.kN??21kkkSSS??5.已知數(shù)列an的前n項(xiàng)和且Sn的最大值為8.21()2nSnknkN?????(1)確定常數(shù)k求an(2)求數(shù)列的前n項(xiàng)和Tn.922nna?解(1)當(dāng)時(shí)取最大值即故從而nkN???212nSnkn???22

5、211822kkk????4k?又所以19(2)2nnnaSSnn??????1172aS??92nan??(2)因?yàn)?9222nnnnanb????1222123112222nnnnnnTbbb???????????????所以21211111222144222222nnnnnnnnnnnTTT????????????????????6.6.設(shè)數(shù)列滿足??na1112(123.......)nnnaaana?????（1）證明對(duì)一切正

6、整數(shù)n成立；21nan??（2）令判斷的大小，并說(shuō)明理由。(123......)nnabnn??1nnbb?與（I）證法一證法一：當(dāng)不等式成立.112211?????an時(shí)22122121.1112232(1)1.12(1)1.kkkkkknkaknkaakkaankak??????????????????????假設(shè)時(shí)成立當(dāng)時(shí)時(shí)時(shí)成立綜上由數(shù)學(xué)歸納法可知，對(duì)一切正整數(shù)成立.12??nan證法二：證法二：當(dāng)n=1時(shí)，.結(jié)論成立.1123