第34講數(shù)列求和之分組轉(zhuǎn)化法與錯(cuò)位相減法求和第1課時(shí)【知識(shí)梳理】1.分組求和法一個(gè)數(shù)列的通項(xiàng)是由的數(shù)列的通項(xiàng)組成的,則求和時(shí)可用分組求和法,分別求和后再相加、減.

2.倒序相加法與并項(xiàng)求和法(1)倒序相加法：如果一個(gè)數(shù)列{an}中,與首末兩端等“距離”的兩項(xiàng)的和相等,那么求這個(gè)數(shù)列的前n項(xiàng)和即可用倒序相加法.(2)并項(xiàng)求和法：數(shù)列{an}滿(mǎn)足彼此相鄰的若干項(xiàng)的和為特殊數(shù)列時(shí),運(yùn)用求其前n項(xiàng)和.如通項(xiàng)公式形如an=(-1)nf(n)的數(shù)列.

3.裂項(xiàng)相消法：把數(shù)列的通項(xiàng)拆成,在求和時(shí)中間的一些項(xiàng)可以相互抵消,從而求得其和.

4.錯(cuò)位相減法：如果一個(gè)數(shù)列的各項(xiàng)是由一個(gè)等差數(shù)列與一個(gè)等比數(shù)列的對(duì)應(yīng)項(xiàng)之構(gòu)成的,那么求這個(gè)數(shù)列的前n項(xiàng)和時(shí)即可用錯(cuò)位相減法.

常用結(jié)論1.一些常見(jiàn)的前n項(xiàng)和公式(1)1+2+3+4+…+n=n(n+1)2.(2)1+3+5+7+…+2n-1=n2.(3)2+4+6+8+…2.常用的裂項(xiàng)公式(1)1n(n+k)=1k(3)1n(n+1)(n+2)=121n(考點(diǎn)01分組、并項(xiàng)求和例1(1)記數(shù)列{an}的前n項(xiàng)和為Sn,若an=(-1)n(2n-1),則S101= ()A.301 B.101C.-101 D.-301(2)已知數(shù)列{an}滿(mǎn)足a1=a2=1,an+2=an+2,n=2k-1,-an,n=2k(k∈(3)數(shù)列{an}(n∈N*)滿(mǎn)足a1=1,前n項(xiàng)和為Sn,對(duì)任意正整數(shù)n都有an+1+an=n+3,則S8= ()A.18 B.28C.40 D.54例2已知數(shù)列an中，a1=1，a(1)求證：數(shù)列an2n(2)求數(shù)列an2n+3例3已知等差數(shù)列an的前n項(xiàng)和為Sn，且S4(1)求數(shù)列an(2)若bn=an?2,n為奇數(shù),2an變式1：已知數(shù)列{an}滿(mǎn)足：a1=1，an+1=an+2n∈(1)求數(shù)列{an}(2)設(shè)cn=an+bn考點(diǎn)02錯(cuò)位相減法求和例4記Sn是各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前n項(xiàng)和,已知S2=34,S4=1516,n∈N(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bnan=2n-1,Tn為數(shù)列{bn}的前n項(xiàng)和,證明:12≤T

例5記數(shù)列an的前n項(xiàng)和為Sn，已知(1)證明：數(shù)列an(2)求數(shù)列an(3)求數(shù)列nan的前n項(xiàng)和變式1：已知數(shù)列an滿(mǎn)足a1=2，an+1=(1)證明：數(shù)列bn(2)求數(shù)列bn的通項(xiàng)公式及前n項(xiàng)和S(3)記dn=bn2n，求數(shù)列變式2：已知等差數(shù)列an的首項(xiàng)a1=3，等比數(shù)列bn的公比q=2，且(1)求數(shù)列an和數(shù)列b(2)設(shè)cn=15anb限時(shí)訓(xùn)練1.若數(shù)列{an}的通項(xiàng)公式是an=(-1)n(3n-2),則a1+a2+…+a10= ()A.15 B.12C.-12 D.-152.在數(shù)列{an}中,已知an+1+an=3·2n,a2=5,則{an}的前11項(xiàng)和為 ()A.2045 B.2046C.4093 D.40943.Sn=12+24+38+…+n2nA.2n-n2n B.2n+1-n4.在數(shù)列{an}中,a1=-40,an+1=an+2,則|a1|+|a2|+…+|a40|= ()A.380 B.800C.880 D.405.已知數(shù)列{an}的前n項(xiàng)和為Sn,且an=n+12n,若Sn≤k恒成立,則k的最小值是 A.2 B.3 C.4 D.56.在數(shù)列{an}中,a1=2,a2=0,且an+2=an+2·(-1)n,則數(shù)列{an}的前32項(xiàng)和S32=()A.128 B.64C.32 D.167.(多選題)在等差數(shù)列{an}中,a2=4,a4+a7=15,設(shè)bn=2an-2+n,則A.an=n+2B.an=2nC.b1+b2+…+b10=2102D.b1+b2+…+b10=21018.數(shù)列{an}滿(mǎn)足an+2-an=2,若a1=1,a4=4,則數(shù)列{an}的前20項(xiàng)和為.

9.若數(shù)列{an}的通項(xiàng)公式為an=13n,則數(shù)列{(n+1)·an}的前n項(xiàng)和Sn=10.已知數(shù)列{an}滿(mǎn)足a1=35,(4an+1)an+1=3an,Sn為數(shù)列1an的前(1)求證:數(shù)列1a(2)求數(shù)列{an}的通項(xiàng)公式;(3)求數(shù)列1an的前n項(xiàng)和S11.(多選題)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,已知a1=1,且3(n+1)an-nan+1=0(n∈N*),則下列結(jié)論正確的是()A.數(shù)列{nan}是等比數(shù)列B.數(shù)列ann是等比數(shù)列C.an=n·3n-1D.Sn12.已知f(x)+f(2-x)=2,an=fn2024(n∈N*),數(shù)列{an}的前n項(xiàng)和為Sn,則S4047= (A.8096 B.8094C.4048 D.404713.已知數(shù)列{an}和數(shù)列{bn},an=2n-1,bn=2-n.設(shè)cn=an·bn,則數(shù)列{cn}的前n項(xiàng)和Sn=.

14.數(shù)列{an}滿(mǎn)足a1+a22+a322+…+an2n-1(1)求{an}的通項(xiàng)公式;(2)若bn=nan,求{bn}的前n項(xiàng)和Tn.參考答案1.A[解析]a1+a2+…+a10=(-1+4)+(-7+10)+…+(-25+28)=3×5=15.2.C[解析]由an+1+an=3·2n,得a1+a2=6,又a2=5,所以a1=1,所以{an}的前11項(xiàng)和為a1+(a2+a3)+(a4+a5)+(a6+a7)+(a8+a9)+(a10+a11)=1+3×(22+24+26+28+210)=4093.故選C.3.B[解析]由Sn=12+24+38+…+n2n,得12Sn=1×122+2×123+3×124+…+n·12n+1,兩式相減得12Sn=12+122+123+124+…+14.B[解析]因?yàn)閍n+1=an+2,所以an+1-an=2,則{an}是首項(xiàng)為-40,公差為2的等差數(shù)列,所以an=a1+(n-1)×2=-40+2n-2=2n-42.當(dāng)n≥21時(shí),an≥0,當(dāng)1≤n≤20時(shí),an<0,所以|a1|+|a2|+…+|a40|=-(a1+…+a20)+(a21+…+a40)=-20×(-40-2)2+5.B[解析]由題知Sn=22+322+423+…+n+12n,所以12Sn=222+323+424+…+n+12n+1,兩式相減可得12Sn=1+122+123+124+…+12n-n+12n+1=1+126.C[解析]在數(shù)列{an}中,a1=2,a2=0,且an+2=an+2·(-1)n,則當(dāng)n為奇數(shù)時(shí),an+2=an-2,所以{an}的奇數(shù)項(xiàng)構(gòu)成首項(xiàng)為2,公差為-2的等差數(shù)列;當(dāng)n為偶數(shù)時(shí),an+2=an+2,所以{an}的偶數(shù)項(xiàng)構(gòu)成首項(xiàng)為0,公差為2的等差數(shù)列.則S32=16×2+16×152×(-2)+16×0+16×152×2=32.7.AD[解析]設(shè)等差數(shù)列{an}的公差為d,由題得a1+d=4,(a1+3d)+(a1+6d)=15,解得a1=3,d=1,所以an=a1+(n-1)·d=n+2,可得bn=2n+n,所以b1+b2+b3+…+b10=(2+1)+(22+2)+(23+3)+…+(210+10)=(2+22+23+…+210)8.210[解析]由題得a2=a4-2=4-2=2,又an+2-an=2,所以數(shù)列{an}的奇數(shù)項(xiàng)構(gòu)成首項(xiàng)為1,公差為2的等差數(shù)列,數(shù)列{an}的偶數(shù)項(xiàng)構(gòu)成首項(xiàng)為2,公差為2的等差數(shù)列,所以數(shù)列{an}的前20項(xiàng)和為a1+a2+…+a20=(a1+a3+…+a19)+(a2+a4+…+a20)=10×1+10×92×2+10×2+10×92×2=9.54-2n+54×3n[解析]因?yàn)?n+1)·an=n+13n,所以Sn=23+332+433+…+n+13n,則13Sn=232+333+…+n3n+n+13n+110.解:(1)證明:由(4an+1)an+1=3an整理得4anan+1+an+1=3an,等式兩邊同時(shí)除以an+1an,可得4+1an=3an+1,則1an由a1=35,可得1a1-2=-13≠0,則數(shù)列1an(2)由(1)得1an-2的通項(xiàng)公式為1an-2=-13n,則1(3)由(2)知1an=2-所以Sn=2-13+2-132+…+2-13n=2n-111.BCD[解析]由3(n+1)an-nan+1=0,可得an+1n+1=3·ann,又a11=1,所以ann為等比數(shù)列,其公比為3,首項(xiàng)為1,所以ann=3n-1,所以an=n·3n-1,故B,C正確;nan=n2·3n-1,因?yàn)?n+1)an+1nan=(n+1)2·3nn2·3n-1=3(n+1)2n2不是常數(shù),所以數(shù)列{nan}不是等比數(shù)列,故A錯(cuò)誤;因?yàn)閍n=n·3n-1,所以Sn=1+2×3+3×32+…+n×3n-1①,所以3Sn=3+2×32+3×33+…+n×3n②,①-②得-12.D[解析]由f(x)+f(2-x)=2,an=fn2024,得當(dāng)1≤n≤2023時(shí),an+a4048-n=fn2024+f4048-n2024=2,即a1+a4047=a2+a4046=…=a2023+a2025=2,又a2024+a2024=f20242024+f20242024=2=a1+a4047,所以2S4047=(a1+a4047)+(a2+a4046)+…+(a4047+a1)=4047(a1+a4047)=8094,所以13.3-2n+32n[解析]∵an=2n-1,bn=2-n,∴cn=an·bn=(2n-1)·2-n,∴Sn=1×2-1+3×2-2+5×2-3+…+(2n-1)×2-n,則12Sn=1×2-2+3×2-3+5×2-4+…+(2n-1)×2-n-1,兩式相減得12Sn=2-1+2(2-2+2-3+2-4+…+2-n)-(2n-1)×2-n-1=12+2×14×1-12n-