www11227007com:高中数学人教B版数学必修5同步课件:2-3-4《数列的综合应用》_图文

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第二章 数 列 第二章 2.3 等比数列 第二章 第4课时 数列的综合应用 课前自主预习 易错疑难辨析 课堂典例讲练 课后强化作业 课前自主预习 情境引入导学 中世纪,意大利数学家斐波那契(1170~1250)在1202年发表 《算盘全书》一书,书中有这样一题:“今有7老妇人共往罗 马,每人有7骡,每骡负7袋,每袋盛有7个面包,每个面包有7把 小刀随之,问列举之物全数共几何?” 知能自主梳理 1.等差数列的通项公式及前n项和公式 数列{an}是等差数列,首项为a1,公差为d,则通项公式an= ____________,前n项和公式Sn=__________=____________. 2.等差数列{an}的主要性质 (1)an=am+________; (2)若m、n、p、q为正整数,m+n=p+q,则 ________________; (3)等差数列{an}的前m项和Sm,则Sm,S2m-Sm,S3m- S2m,?,成________数列. 3.等比数列的通项公式及前n项和公式 首项为 a1,公比为q,则通项公式an=____________,前n项 和公式Sn=__________________. 4.等比数列{an}的主要性质 (1)an=am· __________; (2)若m、n、p、q为正整数,m+n=p+q,则__________. (3)等比数列{an}的前m项和为Sm,则当q≠-1或m为奇数时, Sm,S2m-Sm,S3m-S2m,?,成________数列. 答案: 1.a1+(n-1)d 2.(1)(n-m)d 3.a1· qn-1 4.(1)qn -m n?a1+an? 1 na1+2n(n-1)d 2 (2)am+an=ap+aq (3)等差 ?na1?q=1? ? ?a1?1-qn? ?q≠1? ? ? 1-q (2)am· an=ap· aq (3)等比 预习效果展示 1.如果数列{an}满足a1,a2-a1,a3-a2,?,an-an-1,?, 是首项为1,公比为2的等比数列,那么an=( A.2n+1-1 C.2n-1 B.2n-1 D.2n+1 ) [答案] B [解析] an-an-1=a1qn 1=2n 1, - - ? ?a2-a1=2 ?a3-a2=22 即? ? ? n-1 ? ?an-an-1=2 相加得an-a1=2+22+?+2n-1=2n-2 故an=a1+2n-2=2n-1. 2.已知数列{an}和Sn满足:Sn+Sm=Sn+m,且a1=1,那么a10 =( ) A.1 C.10 [答案] A B.9 D.55 [解析] ∵Sn+Sm=Sn+m,且a1=1,∴S1=1. 可令m=1,得Sn+1=Sn+1, ∴Sn+1-Sn=1. 即当n≥1时,an+1=1,∴a10=1. 3.设等差数列{an}的前n项和为Sn,若a6=S3=12,则数列的 通项公式an=________. [答案] [解析] 2n 设等差数列{an}的首项为a1,公差为d,则 , ?a1+5d=12 ? 则? 3×2 3a + 2 d=12 ? ? 1 ? ?a1+5d=12 即? ? ?a1+d=4 ? ?a1=2 ,解得? ? ?d=2 . ∴an=a1+(n-1)d=2n. 1 1 1 1 1 4. - + - +?- =________. 4 8 16 32 1 024 171 [答案] 1 024 1 1 1 1 1 原式= 2- 3+ 4- 5+?- 10 2 2 2 2 2 [解析] 1 19 4[1-?-2? ] 171 = 1 =1 024. 1-?-2? 课堂典例讲练 等差、等比数列的综合应用 已知{an}是一个公差大于0的等差数列,且满足 a3a6=55,a2+a7=16. (1)求数列{an}的通项公式; b1 b2 b3 (2)若数列{an}和数列{bn}满足等式:an= 2 + 22 + 23 bn +? n(n为正整数),求数列{bn}的前n项和Sn. 2 [解析] 设d>0. (1)解法一:设等差数列{an}的公差为d,则依题 由a2+a7=16,得2a1+7d=16.① 由a3· a6=55,得(a1+2d)(a1+5d)=55.② 由①得2a1=16-7d,将其代入②得(16-3d)(16+3d)= 220,即256-9d2=220, ∴d2=4.又d>0,∴d=2.代入①得a1=1. ∴an=1+(n-1)· 2=2n-1. 解法二:由等差数列的性质得:a2+a7=a3+a6, ? ?a3a6=55 ∴? ? ?a3+a6=16 , 由韦达定理知a3,a6是方程x2-16x+55=0的根,解方程 得x=5或x=11. a6-a3 设公差为d,则由a6=a3+3d,得d= 3 . 11-5 ∵d>0,∴a3=5,a6=11,d= 3 =2, a1=a3-2d=5-4=1. 故an=2n-1. b1 (2)解法一:当n=1时,a1= 2 ,∴b1=2. bn-1 bn b1 b2 b3 当n≥2时,an= 2 +22+23+?+ n-1+2n, 2 bn-1 b1 b2 b3 an-1= 2 +22+23+?+ n-1, 2 bn 两式相减得an-an-1= n,∴bn=2n+1. 2 ? ?2 因此bn=? n+1 ? ?2 ?n=1? . ?n≥2? 当n=1时,S1=b1=2; 当n≥2时,Sn=b1+b2+b3+?+bn b2?1-2n-1? n+2 =2+ =2 -6. 1-2 ∵当n=1时上式也成立, ∴当n为正整数时都有Sn=2n+2-6. bn 解法二:令cn=2n,则有an=c1+c2+?+cn, an+1=c1+c2+?+cn+1, 两式相减得an+1-an=cn+1. 由(1)得a1=1

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