2010届高考数学复习强化双基系列课件__《数列通项的求法》

2010届高考数学复习强化双基系列课件《数列通项的求法》一、公式法二、迭加法若an+1=an+f(n),则:若an+1=f(n)an,则:三、叠乘法an=S1(n=1),Sn-Sn-1(n≥2).an=a1+(ak-ak-1)=a1+f(k-1)=a1+f(k).n-1k=1nk=2nk=2an=a1…=a1f(1)f(2)…f(n-1)(n≥2).anan-1a2a1a3a2四、化归法通过恰当的恒等变形,如配方、因式分解、取对数、取倒数等,转化为等比数列或等差数列.(1)若an+1=pan+q,则:an+1-=p(an-).(3)若an+1=pan+q(n),则:(2)若an+1=,则:panr+qanan+11an1=·+.prpq(4)若an+1=panq,则:lgan+1=qlgan+lgp.五、归纳法先计算数列的前若干项,通过观察规律,猜想通项公式,进而用数学归纳法证之.例已知数列{an}满足:a1=1,an+1=2an+3×2n-1,求{an}的通项公式.an=(3n-1)×2n-2an+1pn+1anpn=+.q(n)pn+11.在数列{an}中,a1=1,Sn=(n≥2),求an.Sn-12Sn-1+1Sn-12Sn-1+1解:由Sn=知:1Sn1Sn-1-=2.1Sn∴{}是以==1为首项,公差为2的等差数列.1S11a11Sn∴=1+2(n-1)=2n-1.∴Sn=.2n-11∵a1=1,当n≥2时,an=Sn-Sn-1=-.(2n-1)(2n-3)2∴an=-,n≥2.1,n=1,(2n-1)(2n-3)2典型例题2.数列{an}的前n项和Sn=n2-7n-8,(1)求{an}的通项公式;(2)求{|an|}的前n项和Tn.解:(1)当n=1时,a1=S1=-14;当n≥2时,an=Sn-Sn-1=2n-8,(2)由(1)知,当n≤4时,an≤0;当n≥5时,an0;当n≥5时,Tn=-S4+Sn-S4=Sn-2S4故an=2n-8,n≥2.-14,n=1,=n2-7n-8-2(-20)∴当n≤4时,Tn=-Sn=-n2+7n+8,=n2-7n+32.故Tn=n2-7n+32,n≥5.-n2+7n+8,n≤4,3.已知数列{an}中,a1=1,an+1=an+1(nN*),求an.12解法一∵an+1=an+1(nN*),12∴an=an-1+1,an-1=an-2+1.1212两式相减得:an-an-1=(an-1-an-2)12∴{an-an-1}是以a2-a1=为首项,公比为的等比数列.1212∴an-an-1=()n-2=()n-1.121212∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=1++()2+…+()n-1121212=2-21-n.即an=2-21-n.解法二由解法一知an-an-1=21-n,又an=an-1+1,12消去an-1得an=2-21-n.解法三∵an=an-1+1,12令an+=(an-1+),12则=-2.∴an-2=(an-1-2).12∴{an-2}是以a1-2=-1为首项,公比为的等比数列.1212∴an-2=-()n-1.即an=2-21-n.3.已知数列{an}中,a1=1,an+1=an+1(nN*),求an.124.数列{an}的前n项和Sn满足条件lgSn+(n-1)lgb=lg(bn+1+n-2),其中,b0且b1.(1)求数列{an}的通项公式;(2)若对nN*,n≥4时,恒有an+1an,试求b的取值范围.解:(1)由已知得lgSnbn-1=lg(bn+1+n-2),∴Snbn-1=bn+1+n-2(b1).∴Sn=b2+(b1).bn-1n-2当n=1时,a1=S1=b2-1;当n≥2时,an=Sn-Sn-1=b2+-b2-bn-1n-2bn-2n-3bn-1(1-b)n+3b-2=.bn-1(1-b)n+3b-2,n≥2.b2-1,n=1,故an=(2)由已知对n≥4恒成立.bn-1(1-b)n+3b-2bn(1-b)(n+1)+3b-2即(n-3)b2-2(n-2)b+(n-1)0对n≥4恒成立.亦即(b-1)[(n-3)b-(n-1)]0对n≥4恒成立.∵b1,∴b对n≥4恒成立.n-3n-1n-3n-1而当n=4时有最大值3,∴b3.5.设Sn是等差数列{an}的前n项和.已知S3与S4的等比中项为S5,S3与S4的等差中项为1,求等差数列{an}的通项an.1513141314解法1:设等差数列{an}的首项a1=a,公差为d,则通项公式为an=a+(n-1)d,前n项和为Sn=na+.n(n-1)d21314依题意有S3S4=(S5)2,(S50)15S3+S4=2,1314由此可得:1314(3a+3d)(4a+6d)=(5a+10d)2,14(3a+3d)+(4a+6d)=2.13251整理得3ad+5d2=0,4a+5d=4.解得d=0,a=1,或a=4.d=-,512∴an=1或an=-n+.512532经验证知an=1时,Sn=5;另一种情况时,Sn=-4,均合题意.∴an=1或an=-n+即为所求数列{an}的通项公式.512532解法2:∵Sn是等差数列的前n项和,故可设Sn=an2+bn,依题意得:1314(a32+b3)(a42+b4)=(a52+b5)2,14(a32+b3)+(a42+b4)=2.13251解得a=0,b=1,或b=.a=-,52665∴Sn=n或Sn=-n2+n.52665在等差数列中,n≥2时,an=Sn-Sn-1,a1亦适合公式.∴an=1或an=-n+.512532整理得13a2+3ab=0,7a+2b=2.5.设Sn是等差数列{an}的前n项和.已知S3与S4的等比中项为S5,S3与S4的等差中项为1,求等差数列{an}的通项an.1513141314解法3:∵Sn是等差数列的前n项和,∴数列{}是等差数列.Snn+=2,S33S55S44依题意得:=()2,S55S33S44+=2,S33S44解得:S4=4,S3=3,S5=5,或S4=,S3=,S5=-4,85524∴a4=S4-S3=1,a5=S5-S4=1或a4=-,a5=-.528516∴an=1或an=-n+.5125325.设Sn是等差数列{an}的前n项和.已知S3与S4的等比中项为S5,S3与S4的等差中项为1,求等差数列{an}的通项an.1513141314解法4:依题意S3=3a2,S4=2(a2+a3),S5=5a3,整理得:3a2+a3=4,a2(a2+a3)=2a32,代入S55S33S44+=2,S33S44=()2,45解得a2=1,a3=1,或a3=-.a2=,85∴an=1或an=-n+.5125325.设Sn是等差数列{an}的前n项和.已知S3与S4的等比中项为S5,S3与S4的等差中项为1,求等差数列{an}的通项an.15131413146.已知an+1=2+an(n∈N+),且a1=a,求an.12解:a1=aa2=2+a12=4-21+2-1a,故猜想:an=4-23-n+21-na,用数学归纳法证明如下:a5=2+a412a3=2+a2=3+a1214=4-20+2-2a,a4=2+a3=+a127218=4-2-1+2-3a,=4-2-2+2-4a,=4-22+20a,证明从略.故an=4-23-n+21-na.解法二:构造等比数列求解(略).7.设数列{an}是公差不为0的等差数列,Sn是数列{an}的前n项和,且S32=9S2,S4=4S2,求数列{an}的通项公式.解:设等差数列{an}的公差为d,由Sn=na1+及已知条件得:n(n-1)d2(3a1+3d)2=9(2a1+d),①4a1+6d=4(2a1+d),②由②得:d=2a1,代入①有:9a12=4a1.解得:a1=0或a1=.49当a1=0时,d=0,与已知条件矛盾,舍去;当a1=时,d=.4989∴an=+(n-1)=n-.49894989故数列{an}的通项公式为an=n-.49898.已知数列{an}是等差数列,且a1=2,a1+a2+a3=12,(1)求数列{an}的通项公式;(2)令bn=an3n,求数列{bn}前n项和的公式.解:(1)设数列{an}的公差为d,则由已知得3a1+3d=12,∴d=2.∴an=2+(n-1)2=2n.故数列{an}的通项公式为an=2n.(2)由bn=an3n=2n3n得数列{bn}前n项和Sn=23+432+…+(2n-2)3n-1+2n3n①∴3Sn=232+433+…+(2n-2)3n+2n3n+1②将①式减②式得:-2Sn=2(3+32+…+3n)-2n3n+1=3(3n-1)-2n3n+1.∴Sn=+n3n+1.3(1-3n)2又a1=2,