高考数学 55数列求和考点专项复习课件 新人教A版 课件VIP免费

[原创]2011届高考数学考点专项复习课件55数列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...