第四节数列求和课时作业练1.(2018江苏泰州期末)设Sn为等差数列{an}的前n项和,S4=14,S10-S7=30,则S9=.答案54解析设等差数列{an}的首项为a1,公差为d,则S4=4a1+6d=14,①S10=10a1+45d,S7=7a1+21d,则S10-S7=3a1+24d=30.②由①②可得d=1,a1=2,故S9=9a1+36d=18+36=54.2.若数列{an}的通项公式为an=2n+2n-1,则该数列的前n项和Sn=.答案2n+1+n2-2解析Sn=(21+1)+(22+3)+(23+5)+…+(2n+2n-1)=(21+22+23+…+2n)+[1+3+5+…+(2n-1)]=2(1-2n)1-2+n[1+(2n-1)]2=2n+1-2+n2=2n+1+n2-2.3.(2018扬州期末检测)若数列{1n(n+1)}的前n项和为Sn,且Sn·Sn+1=34,则正整数n的值为.答案6解析Sn=1-12+12-13+…+1n-1n+1=nn+1,则Sn·Sn+1=nn+1·n+1n+2=nn+2=34,得n=6.4.(2018江苏南通高三调研)数列{an}满足an+an+1=12(n∈N*),a2=2,Sn是数列{an}的前n项和,则S21=.答案721解析 a2=2,a1+a2=12,∴a1=12-2=-32. an+an+1=12,∴a2+a3=a4+a5=a6+a7=…=a20+a21=12,∴S21=a1+(a2+a3)+(a4+a5)+…+(a20+a21)=-32+10×12=72.5.(2018南京、盐城高三第一次模拟)设Sn为等差数列{an}的前n项和,若{an}的前2017项中的奇数项和为2018,则S2017的值为.答案4034解析因为a1+a3+a5+…+a2017=1009a1009,所以a1009=2018÷1009=2,所以S2017=a1+a2+…+a2017=2017a1009=4034.6.(2017徐州王杰中学高三月考)设正项等比数列{an}的首项a1=2,前n项和为Sn,且满足2a3+S2=4,则满足66650,∴q=12,由题意有6665<2(1-q2n)1-q2(1-qn)1-q<1615,∴6665<1+(12)n<1615,∴满足题意的最大正整数n的值为6.7.若数列{an}是首项和公差都为1的等差数列,则数列{1an(an+2)}的前n项和为.答案12(32-1n+1-1n+2)2解析由题意可知an=n,则1an(an+2)=1n(n+2)=12(1n-1n+2),所以所求前n项和为121-13+12-14+13-15+…+1n-1n+2=121+12-1n+1-1n+2=1232-1n+1-1n+2.8.(2018泰州中学高三月度检测)设等比数列{an}满足a1=√2,an=cosθn+√3sinθn,其中θn∈(0,π2),n∈N*,则数列{θn}的前2018项之和是.答案10096π解析因为θn∈(0,π2),所以an=cosθn+√3sinθn=2sin(θn+π6)∈(1,2],设等比数列{an}的公比为q,则q>0,a1=√2,若q>1,则当n充分大时,an>2,矛盾,舍去;若00), 1an+1an+1=32n(n∈N*),∴{1a1+1a2=32,1a2+1a3=34,∴{1a1(1+1q)=32,1a1q(1+1q)=34.∴a1=1,q=2,∴an=2n-1.(2) bn=4n-1+(n-1),∴Sn=(1+0)+(41+1)+(42+2)+…+[4n-1+(n-1)]=(1+41+42+…+4n-1)+[0+1+2+…+(n-1)]=4n-13+(n-1)n2=22n+1+3n2-3n-26.11.在等差数列{an}中,已知公差d=2,a2是a1与a4的等比中项.(1)求数列{an}的通项公式;(2)设bn=an(n+1)2,记Tn=-b1+b2-b3+b4-…+(-1)nbn,求Tn.解析(1)由题意知(a1+d)2=a1(a1+3d),即(a1+2)2=a1(a1+6),解得a1=2,所以数列{an}的通项公式为an=2n.(2)由题意知bn=an(n+1)2=n(n+1),所以Tn=-1×2+2×3-3×4+…+(-1)nn(n+1).4因为bn+1-bn=2(n+1),所以当n为偶数时,Tn=(-b1+b2)+(-b3+b4)+…+(-bn-1+bn)=4+8+12+…+2n=n2(4+2n)2=n(n+2)2;当n为奇数时,Tn=Tn-1+(-bn)=(n-1)(n+1)2-n(n+1)=-(n+1)22.所以Tn={-(n+1)22,n,为奇数n(n+2)2,n.为偶数12.(2018江苏海安高级中学高三月考)已知数列{an}中,首项a1=1,a2=a,an+1=k(an+an+2)对任意正整数n都成立,数列{an}的前n项和为Sn.(1)若k=12,且S18=171,求实数a的值;(2)是否存在实数k,使数列{an}是公比不为1的等比数列,...
