专题13数列的求和一、基础过关题1.数列112,314,518,7116,…,(2n-1)+12n,…的前n项和Sn的值等于()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n【答案】A【解析】该数列的通项公式为an=(2n-1)+12n,则Sn=[1+3+5+…+(2n-1)]+(12+122+…+12n)=n2+1-12n.2.(2016·西安模拟)设等比数列{an}的前n项和为Sn,已知a1=2016,且an+2an+1+an+2=0(n∈N*),则S2016等于()A.0B.2016C.2015D.2014【答案】A3.等差数列{an}的通项公式为an=2n+1,其前n项和为Sn,则数列Snn的前10项的和为()A.120B.70C.75D.100【答案】C【解析】因为Snn=n+2,所以Snn的前10项和为10×3+10×92=75.4.设数列{an}的通项公式为an=2n-7,则|a1|+|a2|+…+|a15|等于()A.153B.210C.135D.120【答案】A【解析】令an=2n-7≥0,解得n≥72.∴从第4项开始大于0,∴|a1|+|a2|+…+|a15|=-a1-a2-a3+a4+a5+…+a15=5+3+1+1+3+…+(2×15-7)=9+1+232=153.5.(2016·福州模拟)已知数列{an}的通项公式为an=1n+1,若前n项和为10,则项数n为________.【答案】1206.在等差数列{an}中,a1>0,a10·a11<0,若此数列的前10项和S10=36,前18项和S18=12,则数列{|an|}的前18项和T18的值是________.【答案】60【解析】由a1>0,a10·a11<0可知d<0,a10>0,a11<0,∴T18=a1+…+a10-a11-…-a18=S10-(S18-S10)=60.7.已知数列{an}中,a1=3,a2=5,且{an-1}是等比数列.(1)求数列{an}的通项公式;(2)若bn=nan,求数列{bn}的前n项和Tn.【答案】(1)an=2n+1.(2)Tn=(n-1)·2n+1+n2+n+42【解析】(1)∵{an-1}是等比数列且a1-1=2,a2-1=4,a2-1a1-1=2,∴an-1=2·2n-1=2n,∴an=2n+1.(2)bn=nan=n·2n+n,故Tn=b1+b2+b3+…+bn=(2+2×22+3×23+…+n·2n)+(1+2+3+…+n).令T=2+2×22+3×23+…+n·2n,则2T=22+2×23+3×24+…+n·2n+1.两式相减,得-T=2+22+23+…+2n-n·2n+1=1-2n1-2-n·2n+1,∴T=2(1-2n)+n·2n+1=2+(n-1)·2n+1.∵1+2+3+…+n=n+12,∴Tn=(n-1)·2n+1+n2+n+42.8.(2016·天津)已知{an}是等比数列,前n项和为Sn(n∈N*),且1a1-1a2=2a3,S6=63.(1)求数列{an}的通项公式;(2)若对任意的n∈N*,bn是log2an和log2an+1的等差中项,求数列{(-1)nb2n}的前2n项和.【答案】(1)an=2n-1.(2)T2n=2n2.9.(2018浙江高考20)已知等比数列{an}的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中项.数列{bn}满足b1=1,数列{(bn+1−bn)an}的前n项和为2n2+n.(Ⅰ)求q的值;(Ⅱ)求数列{bn}的通项公式.【解析】(Ⅰ)由是的等差中项得,所以,解得.由得,因为,所以.点评.本题主要考查等差数列、等比数列、数列求和等基础知识,同时考查运算求解能力和综合应用能力。二、能力提高题1.在数列{an}中,若an+1+(-1)nan=2n-1,则数列{an}的前12项和等于()A.76B.78C.80D.82【答案】B【解析】由已知an+1+(-1)nan=2n-1,得an+2+(-1)n+1·an+1=2n+1,得an+2+an=(-1)n(2n-1)+(2n+1),取n=1,5,9及n=2,6,10,结果相加可得S12=a1+a2+a3+a4+…+a11+a12=78.故选B.2.已知函数f(n)=当n为奇数时,,且an=f(n)+f(n+1),则a1+a2+a3+…+a100等于()A.0B.100C.-100D.10200【答案】B3.(2016·大连模拟)若已知数列的前四项是112+2,122+4,132+6,142+8,则数列的前n项和为______________.【答案】34-2n+3n+2【解析】由前四项知数列{an}的通项公式为an=1n2+2n,由1n2+2n=12(1n-1n+2)知,Sn=a1+a2+a3+…+an-1+an=12[1-13+12-14+13-15+…+(1n-2-1n)+(1n-1-1n+1)+(1n-1n+2)]=12[1+12-1n+1-1n+2]=34-2n+3n+2.4.已知正项数列{an}的前n项和为Sn,∀n∈N*,2Sn=a2n+an.令bn=1an,设{bn}的前n项和为Tn,则在T1,T2,T3,…,T100中有理数的个数为________.【答案】95.若数列{an}的前n项和为Sn,点(an,Sn)在y=16-13x的图象上(n∈N*).(1)求数列{an}的通项公式;(2)若c1=0,且对任意正整数n都有cn+1-cn=.求证:对任意正整数n≥2,总有13≤1c2+1c3+1c4+…+1cn<34.【答案】(1)an=122n+1.(2)见解析(1)解∵Sn=16-13an,∴当n≥2时,an=Sn-Sn-1=13an-1-13an,∴an=14an-1.又∵S1=a1=16-13a1,∴a1=18,∴an=1814n-1=122n+1.∴原式得证.