第八讲数列求和及简单应用1.数列{an}的前n项和Sn=n2+3n+2,则{an}的通项公式为()A.an=2n-2B.an=2n+2C.an={6,n=12n-2,n≥2D.an={6,n=12n+2,n≥22.已知数列{an}满足:an+1=an-an-1(n≥2,n∈N*),a1=1,a2=2,Sn为数列{an}的前n项和,则S2018=()A.3B.2C.1D.03.数列{an}满足an+1+(-1)nan=2n-1,则{an}的前60项和为()A.3690B.3660C.1845D.18304.(2018河南郑州质量预测)已知数列{an}的前n项和为Sn,a1=1,a2=2,且an+2-2an+1+an=0(n∈N*),记Tn=1S1+1S2+…+1Sn(n∈N*),则T2018=()A.40342018B.20172018C.40362019D.201820195.已知数列{an}满足a1=1,a2=3,an+1an-1=an(n≥2),则数列{an}的前40项和S40等于()A.20B.40C.60D.806.已知在数列{an}中,a1=-60,an+1=an+3,则|a1|+|a2|+|a3|+…+|a30|等于()A.445B.765C.1080D.31057.(2018广东惠州模拟)数列{an}的前n项和为Sn,若Sn=2an-2,则数列{nan}的前5项和为.8.已知数列{an}中,a1=1,Sn为数列{an}的前n项和,且当n≥2时,有2ananSn-Sn2=1成立,则S2017=.9.设数列{an}满足:a1=1,a2=3,且2nan=(n-1)an-1+(n+1)·an+1(n≥2),则a20的值是.10.设数列{an}的前n项和为Sn,且a1=1,an+an+1=12n(n=1,2,3,…),则S2n+3=.11.(2018福建福州模拟)已知数列{an}的前n项和为Sn,且Sn=2an-1.(1)证明:数列{an}是等比数列;(2)设bn=(2n-1)an,求数列{bn}的前n项和Tn.112.(2018河北唐山五校联考)已知数列{an}满足:1a1+2a2+…+nan=38(32n-1),n∈N*.(1)求数列{an}的通项公式;(2)设bn=log3ann,求1b1b2+1b2b3+…+1bnbn+1.13.(2018吉林长春质量检测)等差数列{an}的前n项和为Sn,数列{bn}是等比数列,满足a1=3,b1=1,b2+S2=10,a5-2b2=a3.(1)求数列{an}和{bn}的通项公式;(2)若cn={2Sn,n,为奇数bn,n,为偶数设数列{cn}的前n项和为Tn,求T2n.214.(2018天津,18,13分)设{an}是等差数列,其前n项和为Sn(n∈N*);{bn}是等比数列,公比大于0,其前n项和为Tn(n∈N*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;(2)若Sn+(T1+T2+…+Tn)=an+4bn,求正整数n的值.答案精解精析1.Da1=S1=6,当n≥2时,an=Sn-Sn-1=2n+2,∴an={6,n=1,2n+2,n≥2.2.A an+1=an-an-1(n≥2,n∈N*),a1=1,a2=2,∴a3=1,a4=-1,a5=-2,a6=-1,a7=1,a8=2,……,故数列{an}是周期为6的周期数列,且每连续6项的和为0,故S2018=336×0+a2017+a2018=a1+a2=3.故选A.3.D不妨令a1=1,则a2=2,a3=a5=a7=…=1,a4=6,a6=10,……,所以当n为奇数时,an=1,当n为偶数时,构成以a2=2为首项,4为公差的等差数列,所以前60项和为S60=30+2×30+30×(30-1)2×4=1830.4.C由an+2-2an+1+an=0(n∈N*),可得an+2+an=2an+1,所以数列{an}为等差数列,公差d=a2-a1=2-1=1,通项公式为an=a1+(n-1)×d=1+n-1=n,则其前n项和Sn=n(a1+an)2=n(n+1)2,所以1Sn=2n(n+1)=2×(1n-1n+1),Tn=1S1+1S2+…+1Sn=2×(11-12+12-13+…+1n-1n+1)3=2×(1-1n+1)=2nn+1,故T2018=2×20182018+1=40362019,故选C.5.C由an+1=anan-1(n≥2),a1=1,a2=3,可得a3=3,a4=1,a5=13,a6=13,a7=1,a8=3,……,这是一个周期为6的数列,一个周期内的6项之和为263,又40=6×6+4,所以S40=6×263+1+3+3+1=60.6.B an+1=an+3,∴an+1-an=3.∴{an}是以-60为首项,3为公差的等差数列.∴an=-60+3(n-1)=3n-63.令an≤0,得n≤21.∴前20项都为负值.∴|a1|+|a2|+|a3|+…+|a30|=-(a1+a2+…+a20)+a21+…+a30=-2S20+S30. Sn=a1+an2n=-123+3n2×n,∴|a1|+|a2|+|a3|+…+|a30|=765.7.答案258解析解法一:当n=1时,a1=S1=2a1-2,得a1=2,当n≥2时,an=Sn-Sn-1=2an-2-(2an-1-2),得an=2an-1,则数列{an}为等比数列,公比为2,an=2n,a1=2也符合,所以an=2n(n∈N*),所以nan=n·2n,由错位相减法求和得前5项和T5=258.解法二:当n=1时,a1=S1=2a1-2,得a1=2,当n≥2时,an=Sn-Sn-1=2an-2-(2an-1-2),得an=2an-1,则数列{an}为等比数列,公比为2,an=2n,a1=2也符合,所以an=2n(n∈N*),得nan=n·2n,故前5项和T5=2+2×22+3×23+4×24+5×25=2+8+24+64+160=258.8.答案11009解析当n≥2时,由2ananSn-Sn2=1得2(Sn-Sn-1)=(Sn-Sn-1)·Sn-Sn2=-SnSn-1,∴2Sn-2Sn-1=1,又2S1=2,∴{2Sn}是以2为首项,1为公差的等差数列,∴2Sn=n+1,故Sn=2n+1,则S2017=11009.49.答案245解析 2nan=(n-1)an-1+(n+1)an+1(n≥2),∴数列{...
