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課時規(guī)范練31　數(shù)列求和 基礎(chǔ)鞏固組 1.數(shù)列1,3,5,7116,…,(2n-1)+12n,…的前n項(xiàng)和Sn的值等于(　　) A.n2+1-12n B.2n2-n+1-12n C.n2+1-12n-1 D.n2-n+1-12n 2.(2018河北衡水中學(xué)金卷十模,3)已知數(shù)列{an}是各項(xiàng)為正數(shù)的等比數(shù)列,點(diǎn)M(2,log2a2),N(5,log2a5)都在直線y=x-1上,則數(shù)列{an}的前n項(xiàng)和為(　　) A.2n-2 B.2n+1-2 C.2n-1 D.2n+1-1 3.(2018山東濰坊二模,4)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,若Sn=-n2-n,則數(shù)列2(n+1)an的前40項(xiàng)的和為(　　) A.3940 B.-3940 C.4041 D.-4041 4.已知函數(shù)f(x)=xa的圖像過點(diǎn)(4,2),令an=1f(n+1)+f(n),n∈N+.記數(shù)列{an}的前n項(xiàng)和為Sn,則S2 018=　　　　　. 5.(2018浙江余姚中學(xué)4月模擬,17)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且S5=30,S10=110. (1)求Sn; (2)記Tn=1S1+1S2+…+1Sn,求Tn. 6.(2018山西晉城月考)已知數(shù)列{an}滿足a1=3,an+1=2an+(-1)n(3n+1). (1)求證:數(shù)列{an+(-1)nn}是等比數(shù)列; (2)求數(shù)列{an}的前10項(xiàng)和S10. 7.(2018山東濰坊一模,17)公差不為0的等差數(shù)列{an}的前n項(xiàng)和為Sn,已知S4=10,且a1,a3,a9成等比數(shù)列. (1)求{an}的通項(xiàng)公式; (2)求數(shù)列an3n的前n項(xiàng)和Tn. 綜合提升組 8.(2018廣東中山期末)等比數(shù)列{an}中,已知對任意自然數(shù)n,a1+a2+a3+…+an=2n-1,則a12+a22+a32+…+an2等于(　　) A.2n-1 B. (3n-1) C. (4n-1) D.以上都不對 9.(2018湖北重點(diǎn)中學(xué)五模)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,a4=4,S5=15,若數(shù)列1anan+1的前m項(xiàng)和為1011,則m=(　　) A.8 B.9 C.10 D.11 10.(2018山東濰坊三模,17)已知數(shù)列{an}的前n項(xiàng)和為Sn,且1,an,Sn成等差數(shù)列. (1)求數(shù)列{an}的通項(xiàng)公式; (2)若數(shù)列{bn}滿足anbn=1+2nan,求數(shù)列{bn}的前n項(xiàng)和Tn. 11.(2018江西上饒三模,17)已知等比數(shù)列{an}的前n項(xiàng)和為Sn,且6Sn=3n+1+a(n∈N+). (1)求a的值及數(shù)列{an}的通項(xiàng)公式; (2)若bn=(3n+1)an,求數(shù)列{an}的前n項(xiàng)和Tn. 創(chuàng)新應(yīng)用組 12.幾位大學(xué)生響應(yīng)國家的創(chuàng)業(yè)號召,開發(fā)了一款應(yīng)用軟件.為激發(fā)大家學(xué)習(xí)數(shù)學(xué)的興趣,他們推出了“解數(shù)學(xué)題獲取軟件激活碼”的活動.這款軟件的激活碼為下面數(shù)學(xué)問題的答案已知數(shù)列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,…,其中第一項(xiàng)是20,接下來的兩項(xiàng)是20,21,再接下來的三項(xiàng)是20,21,22,依此類推.求滿足如下條件的最小整數(shù)N:N>100且該數(shù)列的前N項(xiàng)和為2的整數(shù)冪.那么該款軟件的激活碼是(　　) A.440 B.330 C.220 D.110 13.(2018云南玉溪月考)數(shù)列{an}滿足:a1=,a2=,且a1a2+a2a3+…+anan+1=na1an+1對任何的正整數(shù)n都成立,則1a1+1a2+…+1a97的值為(　　) A.5 032 B.5 044 C.5 048 D.5 050 參考答案 課時規(guī)范練31　數(shù)列求和 1.A　該數(shù)列的通項(xiàng)公式為an=(2n-1)+12n,則Sn=[1+3+5+…+(2n-1)]+12+122+…+12n=n2+1-12n. 2.C　由題意log2a2=2-1=1,可得a2=2,log2a5=5-1=4,可得a5=16,a5a2=q3=8?q=2,a1=1?Sn=1-2n1-2=2n-1,故選C. 3.D　∵Sn=-n2-n,∴a1=S1=-2. 當(dāng)n≥2時,an=Sn-Sn-1=-n2-n+(n-1)2+(n-1)=-2n, 則數(shù)列{an}的通項(xiàng)公式為an=-2n, 2(n+1)an=2-2n(n+1)=-1n-1n+1, 數(shù)列2(n+1)an的前40項(xiàng)的和為 S40=-1-12+12-13+…+140-141=-4041. 4.2 019-1　由f(4)=2,可得4a=2,解得a=,則f(x)=x12. ∴an=1f(n+1)+f(n)=1n+1+n=n+1-n, S2 018=a1+a2+a3+…+a2 018=(2-1)+(3-2)+(4-3)+…+(2 019-2 018)=2 019-1. 5.解 (1)設(shè){an}的首項(xiàng)為a1,公差為d,由題意得S5=5a1+10d=30,S10=10a1+45d=110,解得a1=2,d=2,所以Sn=n2+n. (2)1Sn=1n(n+1)=1n-1n+1, 所以Tn=1-12+12-13+…+1n-1n+1=1-1n+1=nn+1. 6.(1)證明 ∵an+1=2an+(-1)n(3n+1), ∴an+1+(-1)n+1(n+1)an+(-1)nn =2an+(-1)n(3n+1)-(-1)n(n+1)an+(-1)nn =2[an+(-1)nn]an+(-1)nn=2. 又a1-1=3-1=2, ∴數(shù)列{an+(-1)nn}是首項(xiàng)為2,公比為2的等比數(shù)列. (2)解 由(1)得an+(-1)nn=22n-1=2n,∴an=2n-(-1)nn, ∴S10=(2+22+…+210)+(1-2)+(3-4)+…+(9-10)=2(1-210)1-2-5=211-7=2 041. 7.解 (1)設(shè){an}的公差為d,由題設(shè)可得,4a1+6d=10,a32=a1a9, ∴4a1+6d=10,(a1+2d)2=a1(a1+8d),解得a1=1,d=1.∴an=n. (2)令cn=n3n,則Tn=c1+c2+…+cn=13+232+333+…+n-13n-1+n3n, ① 13Tn=132+233+…+n-13n+n3n+1, ② ①-②得:23Tn=13+132+…+13n-n3n+1 =13(1-13n)1-13-n3n+1=12-123n-n3n+1,∴Tn=34-2n+343n. 8.C　當(dāng)n=1時,a1=21-1=1, 當(dāng)n≥2時,a1+a2+a3+…+an=2n-1,a1+a2+a3+…+an-1=2n-1-1, 兩式做差可得an=2n-2n-1=2n-1,且n=1時,21-1=20=1=a1,∴an=2n-1,故an2=4n-1, ∴a12+a22+a32+…+an2=1(1-4n)1-4=13(4n-1). 9.C　Sn為等差數(shù)列{an}的前n項(xiàng)和,設(shè)公差為d,則a4=4,S5=15=5a3,解得d=1,則an=4+(n-4)1=n. 由于1anan+1=1n(n+1)=1n-1n+1, 則Sm=1-12+12-13+…+1m-1m+1=1-1m+1=1011,解得m=10. 10.解 (1)由已知1,an,Sn成等差數(shù)列,得2an=1+Sn, ① 當(dāng)n=1時,2a1=1+S1=1+a1,∴a1=1. 當(dāng)n≥2時,2an-1=1+Sn-1, ② 由①-②,得2an-2an-1=an, ∴anan-1=2, ∴數(shù)列{an}是以1為首項(xiàng),2為公比的等比數(shù)列, ∴an=a1qn-1=12n-1=2n-1. (2)由anbn=1+2nan得bn=1an+2n, ∴Tn=b1+b2+…+bn =1a1+2+1a2+4+…+1an+2n =1a1+1a2+…+1an+(2+4+…+2n) =1-12n1-12+(2+2n)n2 =n2+n+2-12n-1. 11.解 (1)∵6Sn=3n+1+a(n∈N+), ∴當(dāng)n=1時,6S1=6a1=9+a; 當(dāng)n≥2時,6an=6(Sn-Sn-1)=23n,即an=3n-1, ∵{an}為等比數(shù)列,∴a1=1,則9+a=6,a=-3, ∴{an}的通項(xiàng)公式為an=3n-1. (2)由(1)得bn=(3n+1)3n-1, ∴Tn=b1+b2+…+bn=430+731+…+(3n+1)3n-1, 3Tn=431+732+…+(3n-2)3n-1+(3n+1)3n, ∴-2Tn=4+32+33+…+3n-(3n+1)3n, ∴Tn=(6n-1)3n+14. 12.A　設(shè)數(shù)列的首項(xiàng)為第1組,接下來兩項(xiàng)為第2組,再接下來三項(xiàng)為第3組,以此類推,設(shè)第n組的項(xiàng)數(shù)為n,則前n組的項(xiàng)數(shù)和為n(1+n)2.第n組的和為1-2n1-2=2n-1,前n組總共的和為2(1-2n)1-2-n=2n+1-2-n. 由題意,N>100,令n(1+n)2>100,得n≥14且n∈N+,即N出現(xiàn)在第13組之后.若要使最小整數(shù)N滿足:N>100且前N項(xiàng)和為2的整數(shù)冪,則SN-Sn(1+n)2應(yīng)與-2-n互為相反數(shù),即2k-1=2+n(k∈N+,n≥14),所以k=log2(n+3),解得n=29,k=5.所以N=29(1+29)2+5=440,故選A. 13.B　∵a1a2+a2a3+…+anan+1=na1an+1, ① a1a2+a2a3+…+anan+1+an+1an+2=(n+1)a1an+2, ② ①-②,得-an+1an+2=na1an+1-(n+1)a1an+2, ∴n+1an+1-nan+2=4,同理得nan-n-1an+1=4, ∴n+1an+1-nan+2=nan-n-1an+1, 整理得2an+1=1an+1an+2, ∴1an是等差數(shù)列. ∵a1=14,a2=15, ∴等差數(shù)列1an的首項(xiàng)為4,公差為1,1an=4+(n-1)1=n+3, ∴1a1+1a2+…+1a97=97(4+100)2=5 044.