2025屆新高考數(shù)學(xué)精準(zhǔn)突破復(fù)習(xí)數(shù)列求和問(wèn)題

數(shù)列求和問(wèn)題2025屆新高考數(shù)學(xué)精準(zhǔn)突破復(fù)習(xí)考情預(yù)覽

明確考向1.[數(shù)列求和](多選題)(2021·新高考Ⅱ卷)設(shè)正整數(shù)n=a0·20+a1·2+…+ak-1·2k-1+ak·2k,其中ai∈{0,1},記ω(n)=a0+a1+…+ak,則(

)A.ω(2n)=ω(n) B.ω(2n+3)=ω(n)+1C.ω(8n+5)=ω(4n+3) D.ω(2n-1)=n√√√解析:對(duì)于A選項(xiàng),ω(n)=a0+a1+…+ak,2n=a0×21+a1×22+…+ak-1×2k+ak×2k+1,所以ω(2n)=a0+a1+…+ak=ω(n),A選項(xiàng)正確;對(duì)于B選項(xiàng),取n=2,2n+3=7=1×20+1×21+1×22,所以ω(7)=3,而2=0×20+1×21,則ω(2)=1,即ω(7)≠ω(2)+1,B選項(xiàng)錯(cuò)誤;對(duì)于C選項(xiàng),8n+5=a0×23+a1×24+…+ak×2k+3+5=1×20+1×22+a0×23+a1×24+…+ak×2k+3,所以ω(8n+5)=2+a0+a1+…+ak,4n+3=a0×22+a1×23+…+ak×2k+2+3=1×20+1×21+a0×22+a1×23+…+ak×2k+2,所以ω(4n+3)=2+a0+a1+…+ak,因此,ω(8n+5)=ω(4n+3),C選項(xiàng)正確;對(duì)于D選項(xiàng),2n-1=20+21+…+2n-1,故ω(2n-1)=n,D選項(xiàng)正確.故選ACD.53.[分組轉(zhuǎn)化法求和](2023·全國(guó)乙卷)記Sn為等差數(shù)列{an}的前n項(xiàng)和,已知a2=11,S10=40.(1)求{an}的通項(xiàng)公式;(2)求數(shù)列{|an|}的前n項(xiàng)和Tn.4.[錯(cuò)位相減法求和](2023·全國(guó)甲卷)記Sn為數(shù)列{an}的前n項(xiàng)和,已知a2=1,2Sn=nan.(1)求{an}的通項(xiàng)公式;解:(1)因?yàn)?Sn=nan,當(dāng)n=1時(shí),2a1=a1,即a1=0,當(dāng)n=3時(shí),2(1+a3)=3a3,即a3=2,當(dāng)n≥2時(shí),2Sn-1=(n-1)an-1,所以2(Sn-Sn-1)=nan-(n-1)an-1=2an,考法聚焦

講練突破熱點(diǎn)一分組轉(zhuǎn)化法求和分組求和的策略(1)若數(shù)列{cn}的通項(xiàng)公式為cn=an±bn,且{an},{bn}為等差數(shù)列或等比數(shù)列,則可采用分組求和法求數(shù)列{cn}的前n項(xiàng)和.(3)若數(shù)列的通項(xiàng)公式中有(-1)n等特征,根據(jù)正號(hào)、負(fù)號(hào)分組求和.典例1

(2023·廣西梧州統(tǒng)考一模)已知Sn為數(shù)列{an}的前n項(xiàng)和,Sn+2=2an.(1)求數(shù)列{an}的通項(xiàng)公式;解:(1)因?yàn)镾n+2=2an,所以當(dāng)n=1時(shí),S1+2=a1+2=2a1,解得a1=2,當(dāng)n≥2時(shí),Sn+2=2an,Sn-1+2=2an-1,所以an=2an-2an-1,即an=2an-1,所以數(shù)列{an}是等比數(shù)列,公比為2,首項(xiàng)為a1=2,所以數(shù)列{an}的通項(xiàng)公式為an=2n,n∈N*.分組轉(zhuǎn)化法求和的關(guān)鍵是將數(shù)列通項(xiàng)轉(zhuǎn)化為若干個(gè)可求和的數(shù)列通項(xiàng)的和或差,常見(jiàn)錯(cuò)誤為不能準(zhǔn)確分組或不分奇數(shù)項(xiàng)與偶數(shù)項(xiàng).熱點(diǎn)訓(xùn)練1

(2023·山東煙臺(tái)模擬)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a4=9,S3=15.(1)求{an}的通項(xiàng)公式;解:(1)設(shè)等差數(shù)列{an}的公差為d,由已知得a1+3d=9,3a1+3d=15,解得a1=3,d=2,所以an=2n+1,n∈N*.(2)保持?jǐn)?shù)列{an}中各項(xiàng)先后順序不變,在ak與ak+1(k=1,2,…)之間插入2k個(gè)1,使它們和原數(shù)列的項(xiàng)構(gòu)成一個(gè)新的數(shù)列{bn},記{bn}的前n項(xiàng)和為T(mén)n,求T100的值.熱點(diǎn)二裂項(xiàng)相消法求和裂項(xiàng)相消法就是把數(shù)列的每一項(xiàng)分解,使得相加后項(xiàng)與項(xiàng)之間能夠相互抵消,但在抵消的過(guò)程中,有的是依次項(xiàng)抵消,有的是間隔項(xiàng)抵消.常見(jiàn)的裂項(xiàng)方式有(1)求{an}的通項(xiàng)公式;裂項(xiàng)相消法的基本思路是將通項(xiàng)拆分,可以產(chǎn)生相互抵消的項(xiàng).需注意抵消后并不一定只剩下首尾兩項(xiàng),也有可能前面剩兩項(xiàng),后面剩兩項(xiàng)或者前面剩幾項(xiàng),后面剩幾項(xiàng).(1)求{an}的通項(xiàng)公式;熱點(diǎn)三錯(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,那么求數(shù)列{an·bn}的前n項(xiàng)和Sn時(shí),可采用錯(cuò)位相減法.用錯(cuò)位相減法求和時(shí),應(yīng)注意:(1)等比數(shù)列的公比為負(fù)數(shù)的情形.(2)在寫(xiě)出“Sn”和“qSn”的表達(dá)式時(shí)應(yīng)特別注意將兩式“錯(cuò)項(xiàng)對(duì)齊”,以便準(zhǔn)確寫(xiě)出“Sn-qSn”的表達(dá)式.(1)求{an}和{bn}的通項(xiàng)公式;錯(cuò)位相減法求和需要兩邊先同時(shí)乘等比數(shù)列的公比再錯(cuò)位相減,常見(jiàn)錯(cuò)誤有符號(hào)錯(cuò)誤或不能準(zhǔn)確“錯(cuò)項(xiàng)對(duì)齊”.熱點(diǎn)訓(xùn)練3

(2023·山東濰坊模擬)已知等比數(shù)列{an}的前n項(xiàng)和為Sn,且a1=2,S3=a3+6.(1)求數(shù)列{an}的通項(xiàng)公式;解:(1)設(shè)等比數(shù)列{an