数列求和技巧进阶篇:并项简化计算裂项求和进阶奇偶项数列的处理含答案.pdf

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1、1数列求和技巧进阶篇数列求和技巧进阶篇题型一:(1)已知an-1+an=4n-2,a1=4求Sn(2)已知an-1+an=2n,a1=1求Sn题型二:(1)an=2n-52n,求Sn.(待定系数法)(2)bn=(-1)nan+an+1anan+1=(-1)n1an+1an+1(3)n+2n(n+1)2n=2(n+1)-nn(n+1)2n=2n-1n+112n=1n2n-1-1(n+1)2n(4)n(n+1)=13n(n+1)(n+2)-(n-1)n(n+1),an=2n-1,bn=anan+1(5)cn=4n24n2-1=4n2-1+14n2-1=1+1212n-1-12n+1数列求和技巧进阶

2、篇:并项简化计算,裂项求和进阶,奇偶项数列的处理2题型三:(1)an=2n(-1)n,求数列an的前n项和(2 2)an=(2n-1)(-1)n,求数列,求数列an的前的前n项和项和(3)an=n,bn=2n,求数列(-1)nanbn的前n项和题型四:已知数列an=2n+1,n为奇数2n,n为偶数(1)求数列an的前20项和T20(2)求数列an的前2n项和T2n.(3)求数列an的前2n-1项和T2n-1.(4)求数列an的前n项和Tn题型一题型一并项求和简化计算一般来说,并项求和的计算量比分组求和要小1已知an=2n-1,若bn=ancos2n3,求数列 bn的前3n+1项和T3n+1.2

3、(2023秋湖南长郡中学校考)已知Sn是数列 an的前n项和,a1=2,a2=3,a3=4,数列an+an+1+an+2是公差为1的等差数列,则S40=.3记Sn,为数列 an的前n项和,已知an-1+an=4n-2,a1=4求Sn34已知数列 an的前n项和为Sn,an+1+(-1)nan=2n-1,则S8=5已知 an的前n项和为Sn,an+2+-1n n+12an=n,S50=600,则a1+a2=.6已知an=2n-1,记bn=-1nSn,求数列 bn的前30项的和T30.7已知an=2n+1,设bn=-1nlog2an,数列 bn的前n项和为Tn,求满足 Tk=20的k的值8已知an

4、=n3-1,若bn=an+12cos2n3,求数列 bn的前18项和T189已知an=22n-1,设b1=1,bn+1=an,n为奇数-bn+n,n为偶数,求数列 bn的前2n项和S2n.4题型二题型二裂项求和差比数列的其它处理方式差比数列的其它处理方式(待定系数法待定系数法)1an=2n-52n,求Sn.2an=2n+13n,求Sn.3an=n22n,求Sn.4an=4n2+14n+133n,求Sn.【裂项相加】:(-1)n例:-1n2n+1n n+1=-1n1n+1n+1,本类模型典型标志在通项中含有(-1)n乘以一个分式.对于bn=(-1)nan+an+1anan+1可以裂项为bn=(-

5、1)nan+an+1anan+1=(-1)n1an+1an+151若an=2n-1,数列 bn满足bn=(-1)n+1nanan+1,bn的前n项和为Tn,求Tn2已知数列 an满足an=3n-1,若bn=-1n 2n2+6n+5log231+an+1log231+an+2,求数列 bn的前n项和Tn3已知bn=2n-1,设cn=(-1)n2n+1bn+1+1bn+1,Tn为数列 cn的前n项和,证明:T2n-16.4已知an=2n-1,求(-1)n+12n+1+2n-2an+1an 的前n项和Tn5已知an=6n+1n+2,若bn=2n+3-1nan,求 bn的前n项和Tn.6【等差数列相邻

6、2两项之积构成的的新数列】例如:n(n+1)=13n(n+1)(n+2)-(n-1)n(n+1)一般式,当公差为k时:kn(kn+k)=13kkn(kn+k)(kn+2k)-(kn-k)kn(kn+k)1已知an=2n-1,bn=anan+1,求数列bn的前n项和Tn2已知an=2n-1,bn=-1nanan+1,求数列bn的前n项和Tn.一次乘指数型:分母为一次函数和指数函数相乘例子:n+2n(n+1)2n=2(n+1)-nn(n+1)2n=2n-1n+112n=1n2n-1-1(n+1)2n一般结构a-1kn+ab+ak-bankn-bk n+1+b=1ankn-b-1ank n+1+b1

7、已知bn=3n,若bncn=4 n+14n2-1nN N*,求数列 cn的前n项和2已知an+1=2n,记(an+1)bn=n+2n2+n,Tn为数列 bn的前n项和,求Tn73已知bn=n2+2n+22n,求证:b112+b223+b334+.+bn-1n-1n+bnn n+12.4已知an=n,设bn=an+42an+1anan+1an+2,证明:b1+b2+bn14.对式子变形后再裂项1已知an=12n-1,设cn=4n2anan+1,求数列 cn的前n项和Tn2已知an=2n+4,记bn=1nan,数列 bn的前n项和为Tn,求Tn.3已知an=1n n+1nN*,若bn=2n+1a2

8、n,求数列 bn的前n项和Tn.84已知an=nn+1,证明:a2a1+a3a2+an+1an2023,求n的最小值.3(2023湖南岳阳统考三模)已知an=3n,若bn=log13an,n为奇数an,n为偶数,求数列 bn的前n项和Tn1数列求和技巧进阶篇数列求和技巧进阶篇题型一:(1)已知an-1+an=4n-2,a1=4求Sn(2)已知an-1+an=2n,a1=1求Sn题型二:(1)an=2n-52n,求Sn.(待定系数法)(2)bn=(-1)nan+an+1anan+1=(-1)n1an+1an+1(3)n+2n(n+1)2n=2(n+1)-nn(n+1)2n=2n-1n+112n=

9、1n2n-1-1(n+1)2n(4)n(n+1)=13n(n+1)(n+2)-(n-1)n(n+1),an=2n-1,bn=anan+1(5)cn=4n24n2-1=4n2-1+14n2-1=1+1212n-1-12n+12题型三:(1)an=2n(-1)n,求数列an的前n项和(2 2)an=(2n-1)(-1)n,求数列,求数列an的前的前n项和项和(3)an=n,bn=2n,求数列(-1)nanbn的前n项和题型四:已知数列an=2n+1,n为奇数2n,n为偶数(1)求数列an的前20项和T20(2)求数列an的前2n项和T2n.(3)求数列an的前2n-1项和T2n-1.(4)求数列a

10、n的前n项和Tn题型一题型一并项求和简化计算一般来说,并项求和的计算量比分组求和要小1已知an=2n-1,若bn=ancos2n3,求数列 bn的前3n+1项和T3n+1.【解析】bn=ancos2n3=2n-1cos2n3,【法一】并项求和b3n-1+b3n+b3n+1=6n-3cos6n-23+6n-1cos6n3+6n+1cos6n+23=6n-3cos-23+6n-1cos2+6n+1cos23化简得b3n-1+b3n+b3n+1=-126n-3+6n-1-126n+1=0,故T3n+1=b1+b2+b3+b4+b5+b6+b7+b3n-1+b3n+b3n+1=b1+0=-12【法二】

11、分组求和T3n+1=b1+b2+b3+b3n-2+b3n-1+b3n+b3n+1T3n+1=1-12+3-12+51+6n-5-12+6n-3-12+6n-11+6n+1-12=n 1+6n-52-12+n 3+6n-32-12+n 5+6n-121+6n+1-12=-3n22+n-3n22+3n2+2n-3n-12=-12,3所以,数列 bn的前3n+1项和T3n+1=-122(2023秋湖南长郡中学校考)已知Sn是数列 an的前n项和,a1=2,a2=3,a3=4,数列an+an+1+an+2是公差为1的等差数列,则S40=.【答案】366【分析】设bn=an+an+1+an+2,易得bn

12、=9+n-11=n+8,再由S40=a1+b2+b5+b38求解.【详解】解:设bn=an+an+1+an+2,由题意知 bn是公差为1的等差数列,则b1=a1+a2+a3=9,故bn=9+n-11=n+8,则b2=b1+1=10,故b2+b5+b38=2+8+5+8+38+8=138+13 2+382=364.于是S40=a1+a2+a3+a4+a5+a6+a7+a38+a39+a40,=a1+b2+b5+b38=2+364=366.3记Sn,为数列 an的前n项和,已知an-1+an=4n-2,a1=4求Sn【答案】Sn=n2+n,当n为偶数时n2+n+2,当n为奇数时【详解】解:an-1

13、+an=4 n-1+2=4n-2,nN N*,n2当n为偶数时,Sn=a1+a2+a3+a4+an-1+an=n26+4n-22=n2+n;当n为奇数时,Sn=a1+a2+a3+a4+a5+an-1+an=4+n-1210+4n-22=n2+n+2综上所述,Sn=n2+n,当n为偶数时n2+n+2,当n为奇数时.4已知数列 an的前n项和为Sn,an+1+(-1)nan=2n-1,则S8=【答案】36【解析】由题意可得n为奇数时,an+1-an=2n-1,an+2+an+1=2n+1,两式相减得an+2+an=2;n为偶数时,an+1+an=2n-1,an+2-an+1=2n+1,两式相加得a

14、n+2+an=4n,故S8=a1+a3+a5+a7+a2+a4+a6+a8=2+2+8+24=36.故答案为:365已知 an的前n项和为Sn,an+2+-1n n+12an=n,S50=600,则a1+a2=.【答案】-12【解析】当n=4k+3,kN N时,则n n+12=4k+32k+2为偶数,n+1n+22=2k+24k+5为偶数,可得an+2+-1n n+12an=a4k+5+a4k+3=4k+3,an+3+-1n+1n+22an+1=a4k+6+a4k+4=4k+4,两式相加可得:a4k+6+a4k+5+a4k+4+a4k+3=8k+7,故S50=a1+a2+.+a50=a1+a2

15、+a3+a4+a5+a6+a7+a8+a9+a10+.+a47+a48+a49+a50=a1+a2+7+15+.+95=a1+a2+12 7+952=a1+a2+612=600,4解得a1+a2=-12.6已知an=2n-1,记bn=-1nSn,求数列 bn的前30项的和T30.【解析】Sn=n(1+2n-1)2=n2,所以bn=(-1)nSn=(-1)nn2,所以T30=-12+22-32+42+-292+302=2-1 1+2+4-3 3+4+30-29 29+30=1+2+3+4+29+30=30(1+30)2=465.7已知an=2n+1,设bn=-1nlog2an,数列 bn的前n项

16、和为Tn,求满足 Tk=20的k的值【解析】bn=-1nlog2an=-1nn+1,b2n-1+b2n=(-1)2n-12n+(-1)2n(2n+1)=1,当k为偶数时,Tk=(b1+b2)+(b3+b4)+(bk-1+bk)=k2,令 Tk=k2=20,得k=40;当k为奇数时,Tk=Tk+1-bk+1=k+12-(k+2)=-k+32,令 Tk=k+32=20,得k=37,所以k=40或37.8已知an=n3-1,若bn=an+12cos2n3,求数列 bn的前18项和T18【解析】bn=an+12cos2n3=n3-1+12cos2n3=19n2cos2n3.因为当kN时,b3k-2+b

17、3k-1+b3k=k-232cos 2k-43+k-132cos 2k-23+k2cos2k=k-518,T18=b1+b2+b3+b3+b4+b5+b16+b17+b18=1-518+2-518+3-518+4-518+5-518+6-518=(1+2+3+4+5+6)-5186=583所以数列 bn的前18项和为583.9已知an=22n-1,设b1=1,bn+1=an,n为奇数-bn+n,n为偶数,求数列 bn的前2n项和S2n.【详解】当n为奇数时,bn+1=an=22n-1;则当n为偶数时,bn+bn+1=n.S2n=b1+b2+b2n=b1+b2n+b2+b3+b4+b5+b2n-

18、2+b2n-1=1+a2n-1+2+4+2n-2=1+24n-3+2+2n-2n-12=24n-3+n2-n+1.题型二题型二裂项求和差比数列的其它处理方式差比数列的其它处理方式(待定系数法待定系数法)51an=2n-52n,求Sn.【答案】an=2n-52n=n+1+2n+1-n+2n=n+2+2n,=22+=-5=2=-9 an=2 n+1-92n+1-2n-92n,Sn=2 n+1-92n+1+14=2n-72n+1+14.2an=2n+13n,求Sn.【答案】an=2n+13n=n-1+3n-1-n+3n=2n+2-33n,=12-3=1=1=2,an=n-1+23n-1-n+23n,

19、Sn=2-n+23n.3an=n22n,求Sn.【答案】an=n+12+n+1+t2n+1-n2+n+t2n=n2+4+n+2+2+t2n,令=14+=02+2+t=0=1=-4t=6 an=n+12-4 n+1+62n+1-n2-4n+62nSn=n+12-4 n+1+62n+1-6=n2-2n+32n+1-6.4an=4n2+14n+133n,求Sn.【答案】an=n+12+n+1+t3n+1-n2+n+t3n=2n2+6+2n+3+3+2t3n2=46+2=143+3+2t=13=2=1t=2,an=2 n+12+n+1+23n+1-2n2+n+23n,Sn=2 n+12+n+1+23n

20、+1-15=2n2+5n+53n+1-15.【裂项相加】:(-1)n例:-1n2n+1n n+1=-1n1n+1n+1,本类模型典型标志在通项中含有(-1)n乘以一个分式.对于bn=(-1)nan+an+1anan+1可以裂项为bn=(-1)nan+an+1anan+1=(-1)n1an+1an+11若an=2n-1,数列 bn满足bn=(-1)n+1nanan+1,bn的前n项和为Tn,求Tn【答案】Tn=141+(-1)n+12n+1.6【详解】由题可得bn=(-1)n+1nanan+1=(-1)n+1n(2n-1)(2n+1)=(-1)n+1412n-1+12n+1,所以Tn=141+1

21、3-1413+15+(-1)n+1412n-1+12n+1=141+(-1)n+12n+1.2已知数列 an满足an=3n-1,若bn=-1n 2n2+6n+5log231+an+1log231+an+2,求数列 bn的前n项和Tn【答案】Tn=-1nn+22-14【分析】bn=-1n1n+12+1n+22,分别在n为偶数和n为奇数的情况下,利用裂项相消法和Tn=Tn+1-bn+1求得结果,综合两种情况可得Tn.【详解】bn=-1n 2n2+6n+5log233n+1log233n+2=-1n 2n2+6n+5n+12n+22=-1n1n+12+1n+22,当 n 为 偶 数 时,Tn=-12

22、2-132+132+142+-142-152+-1n2-1n+12+1n+12+1n+22=1n+22-14;当n为奇数时,Tn=Tn+1-bn+1=1n+32-14-1n+22-1n+32=-1n+22-14;综上所述:Tn=-1nn+22-14.3已知bn=2n-1,设cn=(-1)n2n+1bn+1+1bn+1,Tn为数列 cn的前n项和,证明:T2n-16.【详解】cn=(-1)n2n+1(bn+1+1)(bn+1)=(-1)n2n+14(n+1)n=(-1)n141n+1n+1,所以T2n=14-11-12+12+13-.12n-1-12n-12n+12n+1=1412n+1-1,由

23、于T2n=1412n+1-1(1nt-1)是递减的,所以T2nT1=1412+1-1=-16.4已知an=2n-1,求(-1)n+12n+1+2n-2an+1an 的前n项和Tn【解析】(-1)n+12n+1+2n-2an+1an=(-1)n+1an+1+anan+1an=(-1)n+11an+1an+1,所以Tn=1a1+1a2-1a2+1a3+-1n+11an+1an+1=1+-1n+1an+1=1+-1n+12n+1-15已知an=6n+1n+2,若bn=2n+3-1nan,求 bn的前n项和Tn.【详解】bn=2n+3-1nan=6-1n1n+1+1n+2,所以Tn=b1+b2+bn=

24、-612+13+613+14+6-1n1n+1+1n+2=-3+6n+2-1n.【等差数列相邻2两项之积构成的的新数列】7例如:n(n+1)=13n(n+1)(n+2)-(n-1)n(n+1)一般式,当公差为k时:kn(kn+k)=13kkn(kn+k)(kn+2k)-(kn-k)kn(kn+k)1已知an=2n-1,bn=anan+1,求数列bn的前n项和Tn【答案】Tn=13n(4n2+6n-1).【分析】对bn裂项,利用裂项相消法计算作答.【详解】当n2时,anan+1an+2-an-1anan+1=anan+1an+2-an-1=6anan+1=6bn,因此,bn=16(anan+1a

25、n+2-an-1anan+1),nk=2bk=16(anan+1an+2-a1a2a3)=16(anan+1an+2-15),则Tn=nk=2bk+b1=16(anan+1an+2-15)+a1a2=16(2n-1)(2n+1)(2n+3)-52+3=13n(4n2+6n-1),b1=3满足上式,所以Tn=13n(4n2+6n-1).2已知an=2n-1,bn=-1nanan+1,求数列bn的前n项和Tn.【答案】Tn=2n2+2n,n=2k,kN*-2n2-2n+1,n=2k-1,kN*【分析】对n分奇偶讨论,当n为偶数时,采用并项法求和,当n为奇数时,Tn=Tn-1-anan+1【详解】当

26、n为偶数时,Tn=-a1a2+a2a3-a3a4+a4a5+-1nanan+1=a2-a1+a3+a4-a3+a5+an-an-1+an+1=4 a2+a4+an=4n23+2n-12=2n n+1当n为奇数时,当n=1时,T1=-3当n3时,Tn=Tn-1-anan+1=4n-123+2n-32-2n-12n+1=-2n2-2n+1经检验,T1也满足上式,所以当n为奇数时,Tn=-2n2-2n+1综上,数列 bn的前n项和Tn=2n2+2n,n=2k,kN*-2n2-2n+1,n=2k-1,kN*一次乘指数型:分母为一次函数和指数函数相乘例子:n+2n(n+1)2n=2(n+1)-nn(n+

27、1)2n=2n-1n+112n=1n2n-1-1(n+1)2n一般结构a-1kn+ab+ak-bankn-bk n+1+b=1ankn-b-1ank n+1+b81已知bn=3n,若bncn=4 n+14n2-1nN N*,求数列 cn的前n项和【详解】由bncn=4 n+14n2-1nN N*,可得cn=4n+43n2n-12n+1=13n-12n-1-13n2n+1,则数列 cn的前n项和为1301-1313+1313-1325+13n-12n-1-13n2n+1=1-13n2n+1.2已知an+1=2n,记(an+1)bn=n+2n2+n,Tn为数列 bn的前n项和,求Tn【解析】因为(

28、an+1)bn=n+2n2+n,所以bn=n+2n2+nan+1=n+2n n+12n=21n2n-1n+12n+1,所以数列 bn的前n项和为:Tn=21121-1222+1222-1323+1n2n-1n+12n+1=21121-1n+12n+1=1-1n+12n3已知bn=n2+2n+22n,求证:b112+b223+b334+.+bn-1n-1n+bnn n+12.【详解】bnn n+1=n2+2n+2n n+12n=2n2+n+n+2n n+12n+1.=212n+1+1n2n-1n+12n+1b112+b223+b334+.+bn-1n-1n+bnn n+1=2122+1121-1

29、222+123+1222-1323+12n+1+1n2n-1n+12n+1.=21221-1n21-12+1121-1n+12n+1=2 1-1n+12-1n+12n+12.另解:bnn n+1=n2+2n+2n n+12n+1=2n+1n2n-n+2n+12n+1b112+b223+b334+bn-1n-1n+bnn n+1=2212-3222+3222-4323+n+1n2n-n+2n+12n+1=2 1-n+2n+12n+12.得证4已知an=n,设bn=an+42an+1anan+1an+2,证明:b1+b2+bn14.9【详解】解:因为bn=an+42an+1anan+1an+2=n

30、+42n+1n(n+1)(n+2)=1n+1n+42n+1n(n+2),bn=12nn(n+1)-12n+1(n+1)(n+2),故b1+b2+bn=1212-12223+12223-12334+12nn(n+1)-12n+1(n+1)(n+2)=14-12n+1(n+1)(n+2)14.对式子变形后再裂项1已知an=12n-1,设cn=4n2anan+1,求数列 cn的前n项和Tn【解析】cn=4n2anan+1=4n22n-12n+1=4n24n2-1=4n2-1+14n2-1=1+12n-12n+1=1+1212n-1-12n+1Tn=c1+c2+c3+cn=n+121-13+13-15

31、+12n-1-12n+1=n+n2n+1.2已知an=2n+4,记bn=1nan,数列 bn的前n项和为Tn,求Tn.【解析】bn=1nan=1n 2n+4=141n-1n+2,Tn=141-13+12-14+13-15+14-16+1n-1n+2=141+12-1n+1-1n+2=38-2n+34(n+1)(n+2).3已知an=1n n+1nN*,若bn=2n+1a2n,求数列 bn的前n项和Tn.【解析】bn=2n+1n2(n+1)2=1n2-1(n+1)2,则Tn=1-122+122-132+1n2-1(n+1)2=1-1(n+1)2=n n+2(n+1)2.4已知an=nn+1,证明

32、:a2a1+a3a2+an+1an0,即可得证.【详解】解:由an=nn+1,则an+1an=(n+1)2n n+2=n(n+2)+1n n+2=1+121n-1n+2.所以a2a1+a3a2+an+1an=n+121-13+12-14+13-15+1n-1-1n+1+1n-1n+2=n+121+12-1n+1-1n+2=n+34-121n+1+1n+2.10因为1n+1+1n+20,所以n+34-121n+1+1n+2n+34,即a2a1+a3a2+an+1an2023,求n的最小值.【分析】解法一:枚举;解法二:分组求和得出 S2k=k k+12+8 4k-13,进而得出 S2k-1=k

33、k+12+24k-83,求解即可得出答案;解法三:分组求和得出S2k-1=k k+12+24k-83,求解即可得出答案.【详解】解法一:S9=a1+a2+a3+a4+a5+a6+a7+a8+a9=a1+a3+a5+a7+a9+a2+a4+a6+a8=1+2+3+4+5+23+25+27+29=15+680=6952023;又an0,则SnSn+1,且S92023S10,所以n的最小值为10.解法二:kN N*时,S2k=a1+a2+a3+a2k12=a1+a3+a5+a2k-1+a2+a4+a6+a2k=1+2+3+k+23+25+27+22k+1=k k+12+8 4k-13,S2k-1=S

34、2k-a2k=k k+12+8 4k-13-22k+1=k k+12+24k-83,所以S9=S25-1=562+245-83=6952023,又an0,则SnSn+1,且S92023S10,所以n的最小值为10.解法三:当kN N*时,S2k-1=a1+a2+a3+a2k-1=a1+a3+a5+a2k-1+a2+a4+a6+a2k-2=1+2+3+k+23+25+27+22k-1=k k+12+81-4k-11-4=k k+12+8 4k-1-13,所以S9=S25-1=562+8 44-13=6952023.又an0,则SnSn+1,且S92023S10,所以n的最小值为10.3(2023湖南岳阳统考三模)已知an=3n,若bn=log13an,n为奇数an,n为偶数,求数列 bn的前n项和Tn【解析】bn=-n,n为奇数3n,n为偶数,当n为偶数时,Tn=b1+b2+bn=b1+b3+bn-1+b2+b4+bn=-1+3+n-1+32+34+3n=-n2 1+n-12+9 1-9n21-9=983n-1-n24;当n为奇数时Tn=Tn+1-bn+1=983n+1-1-n+124-3n+1=183n+1-98-n+124;综上所述:Tn=183n+1-98-n+124,n为奇数983n-1-n24,n为偶数.

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