2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專題4 數(shù)列 第2講 數(shù)列求和及綜合應(yīng)用 理求數(shù)列的通項(xiàng)訓(xùn)練提示:求數(shù)列通項(xiàng)的常用方法有累加法、累積法、構(gòu)造等比數(shù)列法或已知Sn與an關(guān)系,求an或利用方程思想聯(lián)立方程組,求出基本量,得出an.解題時(shí)應(yīng)注意各自的適用范圍及注意驗(yàn)證n=1的情況.1.(xx寧夏石嘴山高三聯(lián)考)已知各項(xiàng)都不相等的等差數(shù)列an的前7項(xiàng)和為70,且a3為a1和a7的等比中項(xiàng).(1)求數(shù)列an的通項(xiàng)公式;(2)若數(shù)列bn滿足bn+1-bn=an(nN*),且b1=2,求數(shù)列（）的前n項(xiàng)和Tn.解:(1)設(shè)等差數(shù)列an的公差為d(d0),則解得所以an=2n+2.(2)因?yàn)閎n+1-bn=an,所以bn-bn-1=an-1=2n(n2,nN*)bn=(bn-bn-1)+(bn-1-bn-2)+(b2-b1)+b1=an-1+an-2+a1+b1=n(n+1).所以=-,所以Tn=1-+-+-=1-=.【教師備用】 (xx東北三校第二次聯(lián)考)已知數(shù)列an的前n項(xiàng)和為Sn,且a1=2,an+1=Sn+2,nN*.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=n·an,求數(shù)列bn的前n項(xiàng)和Tn.解:(1)當(dāng)n=1時(shí)a2=S1+2=4=2a1,當(dāng)n2時(shí),an+1=2an,數(shù)列an滿足an+1=2an(nN*),且a1=2,所以an=2n(nN*).(2)bn=n·an=n·2nTn=1×21+2×22+3×23+(n-1)·2n-1+n·2n2Tn=1×22+2×23+3×24+(n-1)·2n+n·2n+1兩式相減,得-Tn=21+22+23+2n-1+2n-n·2n+1-Tn=-n·2n+1,Tn=2+(n-1)·2n+1(nN*).求數(shù)列的前n項(xiàng)和訓(xùn)練提示:在數(shù)列求和的幾種常見方法中,一定要注意其各自的適用范圍,其中在裂項(xiàng)相消法中注意裂項(xiàng)后的恒等變形,在錯(cuò)位相減法中注意相減后,哪些項(xiàng)構(gòu)成等比數(shù)列.2.(xx甘肅二診)已知數(shù)列an中,a1=2,且an=2an-1-n+2(n2,nN*).(1)求a2,a3,并證明an-n是等比數(shù)列;(2)設(shè)bn=,求數(shù)列bn的前n項(xiàng)和Sn.解:(1)由已知an=2an-1-n+2(n2,nN*)得a2=4,a3=7.an-n=2an-1-2n+2,即an-n=2an-1-(n-1),因?yàn)?2(n2,nN*).所以an-n是以2為公比的等比數(shù)列.(2)由(1)得an-n=(a1-1)·2n-1.即an=2n-1+n.所以bn=1+.設(shè)cn=,且前n項(xiàng)和為Tn,所以Tn=+Tn=+ -得Tn=1+（+）-=-=2-.所以Tn=4-,Sn=n+4-.【教師備用】 (xx鄭州第二次質(zhì)量預(yù)測(cè))已知等差數(shù)列an的各項(xiàng)均為正數(shù),a1=1,且a3,a4+,a11成等比數(shù)列.(1)求an的通項(xiàng)公式;(2)設(shè)bn=,求數(shù)列bn的前n項(xiàng)和Tn.解:(1)設(shè)等差數(shù)列公差為d,由題意知d>0.因?yàn)閍3,a4+,a11成等比數(shù)列,所以(a4+)2=a3a11,所以(+3d)2=(1+2d)(1+10d),即44d2-36d-45=0,所以d=(d=-舍去),所以an=.(2)bn=(-).所以Tn=(-+-+-)=.類型一:周期數(shù)列與通項(xiàng)公式1.(xx山西大同三模)在數(shù)列an中,已知a1=2,a2=7,an+2等于anan+1(nN+)的個(gè)位數(shù),則axx=. 解析:a1a2=2×7=14,所以a3=4,4×7=28,所以a4=8,4×8=32,所以a5=2,2×8=16,所以a6=6,a7=2,a8=2,a9=4,a10=8,a11=2,所以從第三項(xiàng)起,an的值成周期排列,周期為6,xx=335×6+5,所以axx=a5=2.答案:22.(xx赤峰市高三統(tǒng)考)數(shù)列an滿足a1=1,a2=3,an+2=an+1-an,nN*,則axx=. 解析:因?yàn)閍1=1,a2=3,an+2=an+1-an,所以a3=2,a4=-1,a5=-3,a6=-2,a7=1,a8=3,所以數(shù)列an是以6為周期的周期數(shù)列.所以axx=a6×335+5=a5=-3.答案:-3類型二:由數(shù)列性質(zhì)解決恒成立問題3.(xx遼寧沈陽一模)已知數(shù)列an,cn滿足條件:a1=1,an+1=2an+1,cn=.(1)求證數(shù)列an+1是等比數(shù)列,并求數(shù)列an的通項(xiàng)公式;(2)求數(shù)列cn的前n項(xiàng)和Tn,并求使得am>對(duì)任意nN+都成立的正整數(shù)m的最小值.解:(1)因?yàn)閍n+1=2an+1,所以an+1+1=2(an+1),因?yàn)閍1=1,a1+1=20,所以數(shù)列an+1是首項(xiàng)為2,公比為2的等比數(shù)列.所以an+1=2×2n-1,所以an=2n-1.(2)因?yàn)閏n=(-),所以Tn=(-+-+-)=(-)=.所以=6+,nN*,所以6+15.所以當(dāng)n=1時(shí),取得最大值15.要使得am>對(duì)任意nN*恒成立,結(jié)合(1)的結(jié)果,只需2m-1>15,由此得m>4.所以正整數(shù)m的最小值是5.【教師備用】 (xx東北三校聯(lián)合二模)已知數(shù)列an前n項(xiàng)和為Sn,滿足Sn=2an-2n(nN*).(1)證明:an+2是等比數(shù)列,并求an的通項(xiàng)公式;(2)數(shù)列bn滿足bn=log2(an+2),Tn為數(shù)列()的前n項(xiàng)和,若Tn<a對(duì)正整數(shù)n都成立,求a的取值范圍.解:(1)由題設(shè)Sn=2an-2n(nN*),Sn-1=2an-1-2(n-1)(n2),兩式相減得an=2an-1+2,即an+2=2(an-1+2),又a1+2=4,所以an+2是以4為首項(xiàng),2為公比的等比數(shù)列.an+2=4×2n-1,an=4×2n-1-2=2n+1-2(n2),又a1=2,所以an=2n+1-2(nN*).(2)因?yàn)閎n=log2(an+2)=log22n+1=n+1,=-,所以Tn=(-)+(-)+(-)=-<,所以a,即a的取值范圍為,+).類型三:數(shù)列的綜合問題4.(xx山西大同三模)已知數(shù)列an的前n項(xiàng)和Sn滿足an+3Sn·Sn-1=0(n2,nN*),a1=,則nan的最小值為. 解析:因?yàn)閍n+3Sn·Sn-1=0(n2,nN*),所以Sn-Sn-1+3Sn·Sn-1=0,因?yàn)閍1=,顯然Sn·Sn-10,化簡(jiǎn)得-=3,可見()是以3為首項(xiàng),3為公差的等差數(shù)列,所以=3+3(n-1)=3n,Sn=,從而nan=n(Sn-Sn-1)=-=(1-)(n2),要使nan最小,則需1-(n2)最小,即n=2時(shí)最小,此時(shí)nan=(1-2)=-(n2),當(dāng)n=1時(shí),nan=,故對(duì)任意的nN*,nan最小為-.答案:-5.(xx濱州模擬)已知數(shù)列an中,a1=9,點(diǎn)(an,an+1)在函數(shù)f(x)=x2+2x的圖象上,其中n為正整數(shù).(1)證明:數(shù)列l(wèi)g(an+1)為等比數(shù)列.(2)令bn=an+1,設(shè)數(shù)列bn的前n項(xiàng)積為Tn,即Tn=(a1+1)(a2+1)(an+1),求lg Tn.(3)在(2)的條件下,記cn=,設(shè)數(shù)列cn的前n項(xiàng)和為Sn,求證:Sn<1.(1)證明:由題意得an+1=+2an,即an+1+1=(an+1)2,對(duì)an+1+1=(an+1)2兩邊取對(duì)數(shù)得lg(an+1+1)=2lg(an+1),因?yàn)閍1=9,所以lg(a1+1)=lg 10=1,所以數(shù)列l(wèi)g(an+1)是以1為首項(xiàng),2為公比的等比數(shù)列.(2)解:由(1)知lg(an+1)=2n-1.lg Tn=lg(a1+1)(a2+1)(an+1)=lg(a1+1)+lg(a2+1)+lg(an+1)=所以lg Tn=2n-1.(3)證明:cn=-,Sn=(-)+(-)+(-)+(-)=1-<1.