培優(yōu)課2數(shù)列的求和問題課件高二下學(xué)期數(shù)學(xué)北師大版選擇性

第一章數(shù)列培優(yōu)課2數(shù)列的求和問題北師大版

數(shù)學(xué)

選擇性必修第二冊目錄索引重難探究·能力素養(yǎng)速提升學(xué)以致用·隨堂檢測促達(dá)標(biāo)重難探究·能力素養(yǎng)速提升探究點(diǎn)一分組法求和【例1】

已知在正項(xiàng)等比數(shù)列{an}中,a1+a2=6,a3+a4=24.(1)求數(shù)列{an}的通項(xiàng)公式;(2)數(shù)列{bn}滿足bn=log2an,求數(shù)列{an+bn}的前n項(xiàng)和.解

(1)設(shè)數(shù)列{an}的公比為q(q>0),∴an=a1·qn-1=2×2n-1=2n.(2)bn=log22n=n,設(shè){an+bn}的前n項(xiàng)和為Sn,則Sn=(a1+b1)+(a2+b2)+…+(an+bn)=(a1+a2+…+an)+(b1+b2+…+bn)=(2+22+…+2n)+(1+2+…+n)規(guī)律方法

1.若數(shù)列{cn}的通項(xiàng)公式為cn=an±bn,且{an},{bn}為等差或等比數(shù)列,可采用分組求和法求數(shù)列{cn}的前n項(xiàng)和.2.若數(shù)列{cn}的通項(xiàng)公式為

其中數(shù)列{an},{bn}是等比數(shù)列或等差數(shù)列,可采用分組求和法求{cn}的前n項(xiàng)和.變式訓(xùn)練1已知在等差數(shù)列{an}中,Sn+2=Sn+2n+3(n∈N+).(1)求an;解

(1)設(shè)等差數(shù)列{an}的公差為d,∵Sn+2=Sn+2n+3(n∈N+),所以Sn+2-Sn=an+1+an+2=2n+3,可得an+2+an+3=2n+5,兩式相減可得2d=2,所以d=1;所以an+1+an+2=an+1+an+2=2n+3,可得an=n.探究點(diǎn)二裂項(xiàng)相消法求和【例2】

已知數(shù)列{an}的前n項(xiàng)和為Sn,滿足S2=2,S4=16,{an+1}是等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若an>0,設(shè)bn=log2(3an+3),求數(shù)列

的前n項(xiàng)和.解

(1)設(shè)等比數(shù)列{an+1}的公比為q,其前n項(xiàng)和為Tn,因?yàn)镾2=2,S4=16,所以T2=4,T4=20,當(dāng)q=-2時(shí),a1=-5,所以an+1=(-4)×(-2)n-1=-(-2)n+1.規(guī)律方法

把數(shù)列的每一項(xiàng)拆成兩項(xiàng)之差,求和時(shí)有些部分可以相互抵消,從而達(dá)到求和的目的.常見的拆項(xiàng)公式:變式訓(xùn)練2在①Sn>0,-a3=4,②數(shù)列

的前3項(xiàng)和為6,③an>0且a1,a2,a4+2成等比數(shù)列這三個(gè)條件中任選一個(gè),補(bǔ)充在下面問題中,并求解.已知Sn是等差數(shù)列{an}的前n項(xiàng)和,a1=1,

.

(1)求an;(2)設(shè),求數(shù)列{bn}的前n項(xiàng)和Tn.解

(1)選條件①:設(shè)等差數(shù)列{an}的公差為d,則由

-a3=4得(a1+d)2-(a1+2d)=4,將a1=1代入,解得d=2或d=-2,因?yàn)镾n>0,所以d=2,所以an=2n-1.選條件②:設(shè)等差數(shù)列{an}的公差為d,選條件③:設(shè)等差數(shù)列{an}的公差為d,則由a1,a2,a4+2成等比數(shù)列得(a1+d)2=a1(a1+3d+2),將a1=1代入得d2-d-2=0,解得d=2或d=-1,因?yàn)閍n>0,所以d=2,所以an=2n-1.探究點(diǎn)三錯(cuò)位相減法求和【例3】

設(shè){an}是公比不為1的等比數(shù)列,a1為a2,a3的等差中項(xiàng).(1)求{an}的公比;(2)若a1=1,求數(shù)列{nan}的前n項(xiàng)和.解

(1)設(shè){an}的公比為q,由題設(shè)得2a1=a2+a3,即2a1=a1q+a1q2.所以q2+q-2=0,解得q=1(舍去),q=-2.故{an}的公比為-2.(2)記Sn為{nan}的前n項(xiàng)和.由(1)及題設(shè)可得,an=(-2)n-1.所以Sn=1+2×(-2)+…+n×(-2)n-1,-2Sn=-2+2×(-2)2+…+(n-1)×(-2)n-1+n×(-2)n.規(guī)律方法

1.一般地,如果數(shù)列{an}是等差數(shù)列,{bn}是等比數(shù)列,求數(shù)列{an·bn}的前n項(xiàng)和時(shí),可采用錯(cuò)位相減法.2.用錯(cuò)位相減法求和時(shí)要注意:(1)要善于識別題目類型,特別是等比數(shù)列公比為負(fù)數(shù)的情形.(2)在寫出“Sn”與“qSn”的表達(dá)式時(shí)應(yīng)特別注意將兩式“錯(cuò)項(xiàng)對齊”,以便于下一步準(zhǔn)確地寫出“Sn-qSn”的表達(dá)式.變式訓(xùn)練3已知數(shù)列{an}的通項(xiàng)公式為an=3n-1,在等差數(shù)列{bn}中,bn>0,且b1+b2+b3=15,又a1+b1,a2+b2,a3+b3成等比數(shù)列.(1)求數(shù)列{anbn}的通項(xiàng)公式;(2)求數(shù)列{anbn}的前n項(xiàng)和Tn.解

(1)∵an=3n-1,∴a1=1,a2=3,a3=9.∵在等差數(shù)列{bn}中,b1+b2+b3=15,∴3b2=15,則b2=5.設(shè)等差數(shù)列{bn}的公差為d,又a1+b1,a2+b2,a3+b3成等比數(shù)列,∴(1+5-d)(9+5+d)=64,解得d=-10或d=2.∵bn>0,∴d=-10應(yīng)舍去,∴d=2,∴b1=3,∴bn=2n+1.故anbn=(2n+1)·3n-1,n∈N+.(2)由(1)知Tn=3×1+5×3+7×32+…+(2n-1)3n-2+(2n+1)3n-1,①3Tn=3×3+5×32+7×33+…+(2n-1)3n-1+(2n+1)3n,②①-②,得-2Tn=3×1+2×3+2×32+2×33+…+2×3n-1-(2n+1)3n=3+2(3+32+33+…+3n-1)-(2n+1)3n=3n-(2n+1)3n=-2n·3n.∴Tn=n·3n,n∈N+.學(xué)以致用·隨堂檢測促達(dá)標(biāo)1234567891011121314151617A級必備知識基礎(chǔ)練181.已知數(shù)列{an}的通項(xiàng)公式an=log3,設(shè)其前n項(xiàng)和為Sn,則使Sn<-4成立的最小正整數(shù)n等于(

)A.83 B.82 C.81 D.80C解析

由題意可得an=log3=log3n-log3(n+1),故Sn=log31-log32+log32-log33+…+log3n-log3(n+1)=-log3(n+1)<-4,即log3(n+1)>4,解得n>34-1=80,所以使Sn<-4成立的最小正整數(shù)n等于81.故選C.123456789101112131415161718A.1 B.2 C.3 D.4B1234567891011121314151617183.[探究點(diǎn)二](多選題)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,且

恒成立,則λ的值不可能是(

)A.1 B.0 C.-1 D.2BC1234567891011121314151617184.[探究點(diǎn)二]設(shè),數(shù)列{an}的前n項(xiàng)和Sn=9,則n=

.

991234567891011121314151617185

000

解析

由題意得S100=a1+a2+…+a99+a100=(a1+a3+a5+…+a99)+(a2+a4+…+a100)=(0+2+4+…+98)+(2+4+6+…+100)=100×50=5

000.1234567891011121314151617186.[探究點(diǎn)三]已知在等差數(shù)列{an}中,a4=0,a1+a2=-5.(1)求{an}的通項(xiàng)公式;(2)若cn=3nan,求數(shù)列{cn}的前n項(xiàng)和Sn.123456789101112131415161718(2)因?yàn)閏n=(n-4)3n,所以Sn=-3×31+(-2)×32+(-1)×33+…+(n-5)×3n-1+(n-4)×3n,所以3Sn=-3×32+(-2)×33+(-1)×34+…+(n-5)×3n+(n-4)×3n+1,兩式相減得-2Sn=-3×31+32+33+34+…+3n-(n-4)×3n+1,所以-2Sn=-12+(3+32+33+34+…+3n)-(n-4)×3n+11234567891011121314151617187.[探究點(diǎn)二]已知數(shù)列

是等比數(shù)列,a1=1且a2,a3+2,a4成等差數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=,求數(shù)列{bn}的前n項(xiàng)和Sn.123456789101112131415161718解

(1)設(shè)數(shù)列

的公比為q,∵a1+1=2,∵2(a3+2)=a2+a4,∴2(2q2+1)=2q-1+2q3-1,∴4q2+2=2q+2q3-2,即4(q2+1)=2q(q2+1),解得q=2.∴an+1=2·2n-1=2n,∴an=2n-1.123456789101112131415161718123456789101112131415161718B級關(guān)鍵能力提升練C1234567891011121314151617189.(多選題)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2an-1,bn=log2an+1,則(

)A.數(shù)列{an}是等比數(shù)列B.an=(-2)n-1ACD12345678910111213141516171810.已知數(shù)列{3n+1}與數(shù)列{4n-1},其中n∈N+.它們的公共項(xiàng)由小到大組成新的數(shù)列{an},則{an}前25項(xiàng)的和為(

)A.3197 B.3480 C.3586 D.3775D解析

數(shù)列{3n+1}(n∈N+)的各項(xiàng)為:4,7,10,13,16,19,22,25,28,31,…,數(shù)列{4n-1}(n∈N+)的各項(xiàng)為:3,7,11,15,19,23,27,31,35,39,…,由題意可知,數(shù)列{an}的各項(xiàng)為:7,19,31,…,所以數(shù)列{an}為等差數(shù)列,且首項(xiàng)為7,公差為19-7=12,因此,數(shù)列{an}的前25項(xiàng)的和為7×25+=3

775.故選D.12345678910111213141516171811.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=2,am+n=aman,則S6=(

)A.12 B.27-1 C.27 D.27-2D123456789101112131415161718AD12345678910111213141516171813.已知數(shù)列{an}的前n項(xiàng)和為Sn,且滿足Sn=2an-1(n∈N+),則數(shù)列{nan}的前n項(xiàng)和Tn為

.

(n-1)2n+112345678910111213141516171814.設(shè)等差數(shù)列{an}滿足a2=5,a6+a8=30,則an=

,數(shù)列

的前n項(xiàng)和為

.

2n+1解析

設(shè)數(shù)列{an}的公差為d.∵a6+a8=30=2a7,∴a7=15,∵a7-a2=5d,a2=5,∴d=2,∴an=a2+(n-2)d=2n+1.12345678910111213141516171815.已知{an}是等比數(shù)列,{bn}是等差數(shù)列,且a1=1,b1=3,a2+b2=7,a3+b3=11.(1)求數(shù)列{an}和{bn}的通項(xiàng)公式;(2)設(shè)cn=,n∈N+,求數(shù)列{cn}的前n項(xiàng)和Tn.12345678910111213141516171812345678910111213141516171812345678910111213141516171816.已知函數(shù)f(x)=(x∈R),正項(xiàng)等比數(shù)列{an}滿足a50=1,則f(lna1)+f(lna2)+…+f(lna99)的值是多少?123456789101112131415161718因?yàn)閿?shù)列{an}是等比數(shù)列,所以a1a99=a2a98=…=a49a51==1,即ln

a1+ln

a99=ln

a2+ln

a98=…=ln

a49+ln

a51=2ln

a50=0.所以f(ln

a1)+f(ln

a99)=f(ln

a2)+f(ln

a98)=…=f(ln

a49)+f(ln

a51)=2f(ln

a50)=1.設(shè)S99=f(ln

a1)+f(ln

a2)+f(ln

a3)+…+f(ln