数列求和7种方法(方法全-例子多)

1、DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD1、DDDDDDDDDS秋气”)na小印)dn2122DDDDDDDDDDna,JS(1qn)aaqn1I1n|1q1q(q1)(q1)3、k1n2k(n1)4、S21n(n1)(2n1)n6k例logx1log32xx2x3xn前n项和例DDDlog31xIlog321logxlog2x332DDDDDDDDDDx(1xn)D-1110Sn1(1J)22n112%2%3xn2设SnD1+2+3+D+nDN*,求Sf(n)n最大值(n32)SDDDDDDDDDDDDnSL(n1)(n2)DDDn2n234n64250

2、50S口f(n)nD(n32)Sn1口64口n34n8当nn8,口nD8DDf(n)max题1.等比数列俗的前门项和Sn=2n.1,则十%+%+十5二，则题2.若1(用-1)福(2甩-1)23-3/+甩111解：原式=日-6答案：32飞二、错位相减法求和其中例这种方法是在推导等比数列的前n项和公式时所用的方法，这种方法主要用于求数列anbn的前nDO,a、nb分别是等差数列和等比数列n3求和：13%5%27%3(2n1)xn解：由题可知，(2n1)xn的通项是等差数列2n1的通项与等比数列xn的通项之积设xSn1x3x25x37x4(2n1)xn.(设制错位)得(1x)S12x2x22x32x

3、42xn(2n1)xnn再利用等比数列的求和公式得：(1x)S12xnXn(2n1)xn(2nHl)xn(2n1)xn(1x)(1x)2例12464求数列,22223解：由题可知，2n*的通项是等差数列2n2n的通项与等比数列的通项之积2n设Sn1S2n22_2_22234-6-2n2223242n22nI得(1222223242n)练习题1已知2=*1，求数歹收的前项和答案：SET-21答案：2H的前n项和为三、反序相加法求和这是推导等差数列的前n项和公式时所用的方法，就是将一个数列倒过来排列皿口)，再把它与原数列相加，就可以得到n个”a).例5求证：C03C15C2(2n1)Cn(n1)2

4、nnnn102DDDDDDDDDDDSC03C15C2(2n1)CnDDDDDDDDDD.DnnnnDDDDDDDDDDS(2n1)Cn(2n1)Cn3C1C0nn00CmCn00nnS(2n1)C0(2n1)C13CnCnQDDD.DD.D+2S(2n2)(C0C1CnCn)2(n1),nnn例6sin21。sin22。sin23。sin288。sin289。nODDSsin21。sin22。sin23。sin288。sin289。ODDD.DDDDDDDDDSsin289。sin288。sin23。sin22。sin21。.因为sinxcos(90。x),sin2xcos2x12S(sin

5、21。cos21。)(sin22。cos22。)(sin289。cos289。)口89口S口44.5八卷=1已知函数（1）证明:-幻=1；(2)求八/QA+f-+f+ybojbed的值.解：（1）先利用指数的相关性质对函数化简，后证明左边=右边（2）利用第（1）小题已经证明的结论可知，两式相加得:1S=-所以2E=练习、求值：l2+102+22+92+32+82+10211DDDDDDDDDDDDDDDDD（1）（【）（）（）（!川“.一.v叫D【u（【DD【uzHuz7（1.收）（.3）丁，【uuej.。（.。（.w）soowsoo。呼u肥口（）/（【）/DDIDDDDDDD,DDDDDDD

6、DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD,DDDDDDDDDDDDDDDDDDDDDDDDDDD（1）41（【）（【）（【）（【zu【）（%“）（7）7叫乙也DDDDDDDDDDDDD7）u（7哈讫）.?）（【S07哈气（172）（17）7皿口,DQU叫（1+与）（l+u）uDDD8vZ【ZF【u（【）%产2JSTZZuD-S皿【心口（【）（【）iVVii【）（TI）SDDDDDDDDDDDDDUUVIVVII（3B）B（y）H（力）（【1）S000UUVZVV00ZUUU!I000uZDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

7、DDDDDDDDDDDD(6)n2J.2(n1)nJ.1(n1)2nn(n1)2nn(n1)2nnn(n1)2n(AnB)(AnC)CB(AnBAnC、Jn-JnI1例1、旧、巧居用ynUTnDODDDnn1vn1.nn(1口)122.31DQOQ、；n、nI1口(v21)(32)(工；nWvn)例10DDDaDDannnHlnDDnHl2bDDDDna,nnbDDnnDOODD口ann.nTSi21、)(1口)nnr22DDDb的前nDOnSn口8(1n-4)口n!18(1(34)1)nB1111口nDDD8nmscos1。cos0。cos1。cos1。cos2。cos88。cos89。si

8、n21。cos0。cos1。cos1。cos2。cos88。cos89。sin1。cosn。cos(nH1)。tan(nH1)。tann。(1口)cos0。cos1。cos1。cos2。DODaODcos88。cos89。sin1。口一-(tan1。Itan0。)(tan20Itan1。)I(tan3。Itan2。)tan89。Itan88。口-(tan89。tan0。)口sin1。1cosl。.otl。口sin1。sin21。口DDDDD答案：加十11练习题2。4-+.+3-54-6(盟+1)(邛+3)DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

9、DDDDDDDDDDDDDDDDDDDDDDDDDDDSn.112求cos1+cos2+cos3+cos178+cos17900D.ODDSDcos1+cos2+cos3+cos178+cos179口cosn0cos(180。n。)SDKcos1+cos179)+(cos2+cos178)+(cos3+cos177)+1+(cos89+cos91叫+cos90D0113列aDa1,a3,a2,aaa,求S.n123nH2nn2002ODDSDaaaa20021232002由a1,a3,a2,aaa可得123nB2nn口口口aaaaaa06k6kB26kIB6k*6k爵6kB6SDaaaaTOC

10、 o 1-5 h z20021232002D(aaa)(aa)(aaa)123678126k6kB26kB5Daaaa1999200020012002Daaaa6k6kH26kIB6k*D514DDDDDDDDDDDDDDDaa9,logalogalogaOD563132310ODDSlogalogalogan31323100000000DmnpqaaaamnpqDDDDDDDDlogMlogNlogMN得aaaS(logaloga)(logaloga)(logaloga)n3131032393536口(loga)(loga)(loga)3110329356口log9log9log9333口1

11、0练习、求和:=(q一+9+卜己一+(1-练习题1设，=-J*1）气加-1），则.二答案：L1）,1=1:练习题2.若.+（-1），则+S等于（）?融为奇）-里缶为偶）解：对前n项和要分奇偶分别解决，即：士答案：A练习题3+2的值是解：并项求和，每两项合并，原式=(100+99)+(98+97)+.+(2+1)=5050.答案：DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDTOC o 1-5 h z的前nDDDDDDDDDDD.115求111111nnnn之和.7n个111“ODDD111999-(10k1)v979k个1k个1口1H11urnnnnn-71n个1口1(1011)1(1021)1(1031)1(10n1)11口-(10110210310n)(1H1!1)n个11J0(10n.1).nD-10n-口81(10n-H0B9n)8116口口口a：ann,求(n1)(aa)的口(n1)(n3)nnn:口(n1)(aa)8(n1)nn(n1)(n3)(n2)(n4)(n2)(n4)(n.3)4)口4)8()n2n4n3n4(n1)(aa)4.(8.(nnn2n4n3n4TOC o 1-5 h znnn13:1口口口口a/SODDnDDDDDS4a2(n1,2,),a1,nnn.n1DDDD