2010届高考数学复习强化双基系列课件__《数列的求和》

2010届高考数学复习强化双基系列课件《数列的求和》数列求和的方法将一个数列拆成若干个简单数列,然后分别求和.将数列相邻的两项(或若干项)并成一项(或一组)得到一个新数列(容易求和).一、拆项求和二、并项求和例求和Sn=1×2+2×3+…+n(n+1).例求和Sn=1-2+3-4+5-6+…+(-1)n+1·n.三、裂项求和将数列的每一项拆(裂开)成两项之差,使得正负项能相互抵消,剩下首尾若干项.n2Sn=-,n为偶数时,,n为奇数时.n+12n(n+1)(n+2)3n+1n例求和Sn=++…+.1×212×31n(n+1)1四、错位求和将数列的每一项都作相同的变换,然后将得到的新数列错动一个位置与原数列的各项相减.例等比数列求和公式的推导.五、倒序求和将数列的倒数第k项(k=1,2,3,…)变为正数第k项,然后将得到的新数列与原数列进行变换(相加、相减等).例等差数列求和公式的推导.典型例题(1)已知an=,求Sn;[n(n+1)]22n+1(2)已知an=,求Sn;(2n-1)(2n+1)(2n)2n2+2nn2+2n+12n2+2n2n+1Sn=(3n+2)·2n-1Sn=3n-2n(公比为的等比数列)23(4)Sn=1·n+2·(n-1)+3·(n-2)+…+n·1;法1Sn=1·n+2·(n-1)+3·(n-2)+…+n·[n-(n-1)]=n(1+2+3+…+n)-[21+32+…+n(n-1)]=n(1+2+3+…+n)-[12+22+…+(n-1)2]-[1+2+…+(n-1)]法2Sn=1·n+2·(n-1)+3·(n-2)+…+n·1=1+(1+2)+(1+2+3)+…+(1+2+3+…+n)而an=1+2+3+…+n=n(n+1).12(5)Sn=3n-1+3n-2·2+3n-3·22+…+2n-1.(3)Sn=Cn+4Cn+7Cn+10Cn+…+(3n+1)Cn;0123nn(n+1)(n+2)6课后练习1.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=2,2.将上题(2)中“bn=an3n”改为“bn=anxn(xR)”,仍求{bn}的前n项和.解:令Sn=b1+b2+…+bn,则由bn=anxn=2nxn得:Sn=2x+4x2+…+(2n-2)xn-1+2nxn①∴xSn=2x2+4x3+…+(2n-2)xn+2nxn+1②当x1时,将①式减②式得:(1-x)Sn=2(x+x2+…+xn)-2nxn+1=-2nxn+1.2x(1-xn)1-x∴Sn=-.2x(1-xn)(1-x)22nxn+11-x当x=1时,Sn=2+4+…+2n=n(n+1);综上所述,Sn=n(n+1),x=1时,2x(1-xn)(1-x)22nxn+11-x-,x1时.3.求和:Sn=1+(1+)+(1++)+…+(1+++…+).121412121412n-1121412n-1解:∵an=1+++…+==2-.1-121-1212n-112n-1∴Sn=2n-(1+++…+)121412n-1=2n-2+.12n-14.求数列{n(n+1)(2n+1)}的前n项和Sn.解:∵通项ak=k(k+1)(2k+1)=2k3+3k2+k,∴Sn=2(13+23+…+n3)+3(12+22+…+n2)+(1+2+…+n)n2(n+1)2=++2n(n+1)2n(n+1)(2n+1)2=.n(n+1)2(n+2)25.数列{an}中,an=++…+,又bn=,求数列{bn}的前n项的和.n+11n+12n+1nanan+12解:∵an=(1+2+…+n)=,n+112n∴bn==8(-).2n2n+12n+11n1∴Sn=8[(1-)+(-)+(-)+…+(-)]1213121314n+11n1=8(1-)n+11n+18n=.6.已知lgx+lgy=a,且Sn=lgxn+lg(xn-1y)+lg(xn-2y2)+…+lgyn,求Sn.解:Sn=lgxn+lg(xn-1y)+lg(xn-2y2)+…+lgyn,又Sn=lgyn+lg(xyn-1)+…+lg(xn-1y)+lgxn,∴2Sn=lg(xnyn)+lg(xnyn)+…+lg(xnyn)+lg(xnyn)n+1项=n(n+1)lg(xy).∵lgx+lgy=a,∴lg(xy)=a.∴Sn=lg(xy)=a.n(n+1)2n(n+1)2注:本题亦可用对数的运算性质求解:∴Sn=lg(xy)=a.n(n+1)2n(n+1)2∵Sn=lg[xn+(n-1)+…+3+2+1y1+2+3+…+(n-1)+n],8.求数列1,2+3,4+5+6,7+8+9+10,…的通项an及前n项和Sn.解:an=[+1]+[+2]+…+[+n]n(n-1)2n(n-1)2n(n-1)2n2(n-1)2=+=n3+n.n(n+1)21212∴Sn=(13+23+…+n3)+(1+2+…+n)1212n(n+1)2=[]2+1212n(n+1)2=(n4+2n3+3n2+2n).187.求证:Cn+3Cn+5Cn+…+(2n+1)Cn=(n+1)2n.012n证:令Sn=Cn+3Cn+5Cn+…+(2n+1)Cn.012n又Sn=(2n+1)Cn+(2n-1)Cn+…+3Cn+Cn,nn-110∴2Sn=2(n+1)(Cn+Cn+…+Cn)=2(n+1)2n.01n∴Cn+3Cn+5Cn+…+(2n+1)Cn=(n+1)2n.012n9.已知递增的等比数列{an}前3项之积为512,且这三项分别减去1,3,9后又成等差数列,求数列{}的前n项和.ann解:设等比数列{an}的公比为q,依题意得:a1a2a3=512a23=512a2=8.∵前三项分别减去1,3,9后又成等差数列,∴(-1)+(8q-9)=2(8-3)q=2或q=(舍去).q812∴an=a2qn-2=82n-2=2n+1.∴所求数列的前n项和Sn=++…+①1222232n+1n2n+1n-1123224∴Sn=++…++②122n+2n①-②得:Sn=++…+-2n+11122123122n+2n∴Sn=++…+-12n1222n+1n12=1--.12n2n+1n10.已知数列{an}中,a1=1,(2n+1)an=(2n-3)an-1(n≥2,nN*),求数列{an}的前n项和Sn.∴=.an-1an2n-32n+1∴Sn=a1+a2+…+an解:∵(2n+1)an=(2n-3)an-1,则=,…,=,=.an-2an-12n-52n-1a2a337a1a215∴=.a1an(2n+1)(2n-1)3∴an=(2n+1)(2n-1)3=(-).3212n-112n+13212n-112n+1=[(1-)+(-)+(-)+…+(-)]13151315173n2n+1=.解:(1)a1C-a2C+a3C=a1-2a1q+a1q2=a1(1-q)2.22221011.已知{an}是首项为a1,公比为q的等比数列.(1)求和:a1C2-a2C2+a3C2,a1C3-a2C3+a3C3-a4C3;(2)由(1)的结果归纳概括出关于正整数n的一个结论,并加以证明;(3)设q≠1,Sn是{an}的前n项和,求S1Cn-S2Cn+S3Cn-S4Cn+…+(-1)nSn+1Cn.00011122233n3210a1C-a2C+a3C-a4C=a1-3a1q+3a1q2-a1q3=a1(1-q)3.3333(2)归纳概括的结论为:a1C-a2C+a3C-a4C+…+(-1)nan+1C=a1(1-q)n,其中,3n210nnnnnn为正整数.证明如下:a1C-a2C+a3C-a4C+…+(-1)nan+1C3n210nnnnn=a1C-a1qC+a1q2C-a1q3C+…+(-1)na1qnC3n210nnnnn=a1[C-qC+q2C-q3C+…+(-1)nqnC]3n210nnnnn=a1(1-q)n.∴a1C-a2C+a3C-a4C+…+(-1)nan+1C=a1(1-q)n.3n210nnnnn解:(3)记t=,则由Sn=t(1-qn)得:1-qa10123nS1Cn-S2Cn+S3Cn-S4Cn+…+(-1)nSn+1Cn=t[(1-q)Cn-(1-q2)Cn+(1-q3)Cn+…+(-1)n(1-qn+1)Cn]012n0123n-tq[Cn-qCn+q2Cn-q3Cn+…+(-1)nqnCn]=t[Cn-Cn+Cn-Cn+…+(-1)nCn]012n3=t(1-1)n-tq(1-q)n=-tq(1-q)n,从而有:0123nS1Cn-S2Cn+S3Cn-S4Cn+…+(-1)nSn+1Cn=-tq(1-q)n=-(1-q)n.1-qa1q11.已知{an}是首项为a1,公比为q的等比数列.(1)求和:a1C2-a2C2+a3C2,a1C3-a2C3+a3C3-a4C3;(2)由(1)的结果归纳概括出关于正整数n的一个结论,并加以证明;(3)设q≠1,Sn是{an}的前n项和,求S1Cn-S2Cn+S3Cn-S4Cn+…+(-1)nSn+1Cn.00011122233n(1)证:由已知S1=a1=a,Sn=aqn-1,当n≥2时,an=Sn-Sn-1=aqn-1-aqn-2=a(q-1)qn-2.∴在{an}中,从第2项开始成等比数列.12.数列{an}中,a1=a,前n项和Sn构成公比为q(q1)的等比数列.(1)求证:在{an}中,从第2项开始成等比数列;(2)当a=250,q=时,设bn=log2|an|,求|b1|+|b2|+…+|bn|.12an+1an∵==q(n≥2),a(q-1)qn-2a(q-1)qn-1(2)解:由(1)知an=a,n=1,a(q-1)qn-2,n≥2.当a=250,q=时,b1=log2|a1|=log2250=50,12n≥2时,bn=log2|an|=log2|250(-1)()n-2|=51-n,1212∴bn=51-n(nN*).∴当1≤n≤51时,|b1|+|b2|+…+|bn|=(51-1)+(51-2)+…+(51-n)=51n-n(n+1)2=-n2+n;101212当n≥52时,|b1|+|b2|+…+|bn|=+50(50+1)2(n-51)(1+n-51)2=n2-n+2550.101212n2-n+2550,n≥52.101212综上所述|b1|+|b2|+…+|bn|=-n2+n,1≤n≤51,101212=(50+49+…+1)+[1+2+…+(n-51)]=51n-(1+2+…+n)