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《等差数列》单元复习课《等差数列》单元复习课标要求:1.通过实例,理解等差数列的概念.2.探索并掌握等差数列的通项公式与前n项和公式.3.能在具体的问题情境中,发现数列的等差关系.4.体会等差数列与一次函数的关系.例.等差数列{na}中,15,3104aa,求通项na及前n项和nS.例题讲解例.等差数列{na}中,15,3104aa,求通项na及前n项和nS.解:由1593311dada,解得231da,故52)1(1ndnaan,nnnnnaaSnn42)523(2)(21.例题讲解1.等差数列{na}中,若11a,2273aa,则10a.2.等差数列{na}的前n项和为nS,已知3321aaa,426504948aaa,求50S.3.等差数列{na}中,12,2051aa,求通项na及前n项和nS的最大值.课堂练习1.等差数列{na}中,若11a,2273aa,则10a.解:由228211731daaaa,解得311da,故269110daa.课堂练习2.等差数列{na}的前n项和为nS,已知3321aaa,426504948aaa,求50S.课堂练习2.等差数列{na}的前n项和为nS,已知3321aaa,426504948aaa,求50S.解:由4263484950321aaaaaa,得429)(3501aa,143501aa所以3575250)(50150aaS.课堂练习3.等差数列{na}中,12,2051aa,求通项na及前n项和nS的最大值.课堂练习3.等差数列{na}中,12,2051aa,求通项na及前n项和nS的最大值.解一:由215,20151aada,得222)1(1ndnaan,nnnaaSnn212)(21,二次函数xxy212开口向下,对称轴为221x,所以当10n或11n时,nS取最大值1101110SS.课堂练习3.等差数列{na}中,12,2051aa,求通项na及前n项和nS的最大值.解二:由215,20151aada,得222)1(1ndnaan,可知数列{na}为单调递减数列,令0na,11n,当11n时,0na,当11n时,0na,所以当10n或11n时,nS取最大值1101110SS.课堂练习1.等差数列{na}中,若11a,2273aa,则5a,9a.2.等差数列{na}的前n项和为nS,若5,252Sa,求nS.3.等差数列{na}的前n项和为nS,若,16,442SS则6S=.巩固练习1.等差数列{na}中,若11a,2273aa,则5a,9a.解:由591732aaaaa,解得239a,115a.巩固练习2.等差数列{na}的前n项和为nS,若5,252Sa,求nS.巩固练习2.等差数列{na}的前n项和为nS,若5,252Sa,求nS.解:由553543215aaaaaaS得13a,故323aad,所以521daa,nnnnndandSn21323)235(23)2(22212巩固练习3.等差数列{na}的前n项和为nS,若,16,442SS则6S=.巩固练习3.等差数列{na}的前n项和为nS,若,16,442SS则6S=.解:由ndandSn)2(212,根据题意16)2(484)2(2211daddad,解得211da,所以36)2(61816dadS.巩固练习3.等差数列{na}的前n项和为nS,若,16,442SS则6S=.思路二:由2S,24SS,46SS成等差数列,得)(2)(24462SSSSS,整理得)(3246SSS所以361236S.巩固练习3.等差数列{na}的前n项和为nS,若,16,442SS则6S=.思路三:由22S,44S,66S成等差数列,得4262462SSS整理得666S,所以366S.巩固练习归纳总结等差数列{na}:定义:当2n时,daann1(常数)通项公式:dnaan)1(1(累加法)等差中项:bAa,,成等差数列,则2baA性质:若qpnm,则qpnmaaaa推广式:dmnaamn)(求公差:mnaadmn归纳总结等差数列的前n项和nS:通项公式:ndandnaaSnn)2(22)(121性质1:mS,mmSS2,mmSS23,成等差数列性质2:由)2(21dandnSn,知nSn为等差数列归纳总结
本文标题:等差数列复习课教学课件
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