2013高考人教a版文科數(shù)學一輪強化訓練5.3等比數(shù)列及其前n項和

1、1.設為等比數(shù)列的前n項和則等于()nSna2580aa????52SSA.11B.8C.5D.11答案:A解析:由∴即.2580aa???582aa???382qq?????∴.5(1)151153311223(1)1211aqSqqSaqqq????????????2.在等比數(shù)列中公比|q|.若則m等于()na11a???1?12345maaaaaa??A.9B.10C.11D.12答案:C解析:.51010123451111maa

2、aaaaaqaqa????3.在公比為整數(shù)的等比數(shù)列中如果那么該數(shù)列的前8項和為na1418aa???2a312a???()A.513B.512C.510D.2258答案:C解析:q=或q=2而Z3211313(1)18()1222qaqaqqqq??????????12q?∴.122qa???∴.9882(12)2251012S??????4.在正項等比數(shù)列中則.na153537225aaaaaa?????35aa??答案:5解

3、析:5.2223355353()2()()25aaaaaaa???????5a?5.等比數(shù)列的前n項和為則數(shù)列的前n項和.na21n??2nanT?答案:413n?解析:∵當時21nnS???2n?1121nnS??????∴12nna???∴214nna???∴.2114aq???∴.1441143nnnT?????6.等比數(shù)列中已知.na14216aa???解析:設公比為q則233(1)631333SqSqSS??????3q?于

4、是.36197124312316SqqSq?????????4.等比數(shù)列的公比q0已知則的前4項和.na221naa????16nnaa???Ana4S?答案:152解析:由得216nnnaaa????116nnnqqq??????即q6=0q0解得q=2.2q?又21a??所以.1441(12)15122122aS??????5.三個數(shù)成等差數(shù)列其比為3∶4∶5如果最小數(shù)加上1則三數(shù)成等比數(shù)列那么原三數(shù)分別為.答案:152025

5、解析:設原三數(shù)為則解得t=5345(0)tttt????2(31)516ttt???∴3t=154t=205t=25.∴原三數(shù)為152025.題組二等比數(shù)列的判斷6.已知等比數(shù)列的通項公式且求證:成等比數(shù)na113()2nna???nb?32313nnnaaa?????nb列.證明:∵113()2nna????∴32313nnnnbaaa?????3332113()3()22nn?????3113()2n?331113()(1)22

6、4n????.33211()42n??∴.311()2bnbn??∴成等比數(shù)列.nb7.設為數(shù)列的前n項和N其中k是常數(shù).nSna2nSknnn???????(1)求及1ana(2)若對于任意的N成等比數(shù)列求k的值.m?24mmmaaa????解:(1)當n=1時111aSk?????當時(n1)]=2knk1.()2n?21[(nnnaSSknnkn????????21)?經(jīng)驗證當n=1時()式成立∴.21naknk???(2)∵