20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：64-數(shù)列求和-【含解析】-文庫(kù)吧
2025-04-03 02:51 本頁(yè)面
【正文】 {2n1}與{3n2}的公共項(xiàng)從小到大排列得到數(shù)列{an},則{an}的前n項(xiàng)和為 .關(guān)鍵能力學(xué)案突破 考點(diǎn)分組求和【例1】(2020陜西榆林一模,理19)已知數(shù)列{an},{bn}滿(mǎn)足a1=1,b1=12,2an+1=an+12bn,2bn+1=12an+bn.(1)證明:數(shù)列{an+bn},{anbn}為等比數(shù)列。(2)記Sn為數(shù)列{an}的前n項(xiàng)和,證明:Sn103.思考具有什么特點(diǎn)的數(shù)列適合分組求和?解題心得具有下列特點(diǎn)的數(shù)列適合分組求和:(1)若an=bn177。,且{bn},{}為等差數(shù)列或等比數(shù)列,可采用分組求和法求{an}的前n項(xiàng)和。(2)通項(xiàng)公式為an=bn,n為奇數(shù),n為偶數(shù)的數(shù)列,其中數(shù)列{bn},{}是等比數(shù)列或等差數(shù)列,可采用分組求和法求和.對(duì)點(diǎn)訓(xùn)練1(2020天津河?xùn)|一模,16)已知遞增等差數(shù)列{an},等比數(shù)列{bn},數(shù)列{},a1=c1=1,c4=9,a1,a2,a5成等比數(shù)列,bn=an+,n∈N*.(1)求{an},{bn}的通項(xiàng)公式。(2)求數(shù)列{}的前n項(xiàng)和Sn.考點(diǎn)并項(xiàng)求和【例2】(2020河南安陽(yáng)二模,理17)記數(shù)列{an}的前n項(xiàng)和為Sn,已知Sn=2an2n+1.(1)求數(shù)列{an}的通項(xiàng)公式。(2)記bn=(1)nlog223(an+4)43,數(shù)列{bn}的前n項(xiàng)和為T(mén)n,求Tn.思考具有什么特點(diǎn)的數(shù)列適合并項(xiàng)求和?解題心得若數(shù)列{an}的通項(xiàng)公式為an=(1)nf(n),則一般利用并項(xiàng)求和法求數(shù)列{an}的前n項(xiàng)和.對(duì)點(diǎn)訓(xùn)練2已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且a1=1,S3+S4=S5.(1)求數(shù)列{an}的通項(xiàng)公式。(2)令bn=(1)n1an,求數(shù)列{bn}的前n項(xiàng)和Tn.考點(diǎn)錯(cuò)位相減法求和【例3】(2020全國(guó)3,理17)設(shè)數(shù)列{an}滿(mǎn)足a1=3,an+1=3an4n.(1)計(jì)算a2,a3,猜想{an}的通項(xiàng)公式并加以證明。(2)求數(shù)列{2nan}的前n項(xiàng)和Sn.思考具有什么特點(diǎn)的數(shù)列適合用錯(cuò)位相減法求和?,如果數(shù)列{an}是等差數(shù)列,{bn}是等比數(shù)列,求數(shù)列{anbn}的前n項(xiàng)和時(shí),可采用錯(cuò)位相減法求和,解題思路是:和式兩邊先同乘等比數(shù)列{bn}的公比,再作差求解.“Sn”與“qSn”的表達(dá)式時(shí),應(yīng)特別注意將兩式“錯(cuò)項(xiàng)對(duì)齊”,以便下一步準(zhǔn)確寫(xiě)出“SnqSn”的表達(dá)式.對(duì)點(diǎn)訓(xùn)練3(2020全國(guó)1,理17)設(shè){an}是公比不為1的等比數(shù)列,a1為a2,a3的等差中項(xiàng).(1)求{an}的公比。(
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