第2课时an与Sn的关系及裂项求和法课后篇巩固探究A组1.已知数列{an}的前n项和Sn=1n,则a5的值等于()A.120B.-120C.130D.-130解析:a5=S5-S4=15−14=-120.答案:B2.已知等差数列{an}的前n项和为Sn,a5=5,S5=15,则数列{1anan+1}的前100项和为()A.100101B.99101C.99100D.101100解析: S5=5(a1+a5)2=5(a1+5)2=15,∴a1=1,∴d=a5-a15-1=5-15-1=1,∴an=1+(n-1)×1=n,∴1anan+1=1n(n+1).设{1anan+1}的前n项和为Tn,则T100=11×2+12×3+…+1100×101=1-12+12−13+…+1100−1101=1-1101=100101.答案:A3.设{an}(n∈N+)是等差数列,Sn是其前n项和,且S5
S8,则下列结论错误的是()A.d<0B.a7=0C.S9>S5D.S6和S7均为Sn的最大值解析:由S50.1又S6=S7,∴a1+a2+…+a6=a1+a2+…+a6+a7,∴a7=0,故B正确;同理由S7>S8,得a8<0,又d=a7-a6<0,故A正确;由C选项中S9>S5,即a6+a7+a8+a9>0,可得2(a7+a8)>0.而由a7=0,a8<0,知2(a7+a8)>0不可能成立,故C错误; S5S8,∴S6与S7均为Sn的最大值,故D正确.故选C.答案:C4.数列{1(n+1)2-1}的前n项和Sn为()A.n+12(n+2)B.34−n+12(n+2)C.34−12(1n+1+1n+2)D.32−1n+1−1n+2解析:1(n+1)2-1=1n2+2n=12(1n-1n+2),于是Sn=12¿1n-1n+2)=34−12(1n+1+1n+2).答案:C5.设函数f(x)满足f(n+1)=2f(n)+n2(n∈N+),且f(1)=2,则f(20)为()A.95B.97C.105D.192解析: f(n+1)=f(n)+n2,∴f(n+1)-f(n)=n2.∴f(2)-f(1)=12,2f(3)-f(2)=22,……f(20)-f(19)=192,∴f(20)-f(1)=1+2+…+192=(1+19)×1922=95.又f(1)=2,∴f(20)=97.答案:B6.已知数列{an}的前n项和Sn=n2-9n,第k项满足50;当n≤3时,an=2n-7<0,∴T5=-(a1+a2+a3)+a4+a5=13.当1≤n≤3时,Tn=-(a1+a2+…+an)=-n2+6n;当n≥4时,Tn=-(a1+a2+a3)+a4+a5+…+an=n2-6n+18.综上所述,Tn={-n2+6n,1≤n≤3,n2-6n+18,n≥4.B组1.若等差数列{an}的通项公式为an=2n+1,则由bn=a1+a2+…+ann所确定的数列{bn}的前n项之和是()A.n(n+2)B.12n(n+4)C.12n(n+5)D.12n(n+6)4解析:由题意知a1+a2+…+an=n(3+2n+1)2=n(n+2),∴bn=n(n+2)n=n+2.于是数列{bn}的前n项和Sn=n(3+n+2)2=12n(n+5).答案:C2.已知一个等差数列共n项,且其前四项之和为21,末四项之和为67,前n项和为286,则项数n为()A.24B.26C.25D.28解析:设该等差数列为{an},由题意,得a1+a2+a3+a4=21,an+an-1+an-2+an-3=67,又a1+an=a2+an-1=a3+an-2=a4+an-3,∴4(a1+an)=21+67=88,∴a1+an=22.∴Sn=n(a1+an)2=11n=286,∴n=26.答案:B3.已知数列{an}满足a1=1,an=an-1+2n(n≥2),则a7=()A.53B.54C.55D.109解析: an=an-1+2n,∴an-an-1=2n.∴a2-a1=4,a3-a2=6,a4-a3=8,…,an-an-1=2n(n≥2).∴an=1+4+6+…+2n=1+(n-1)(4+2n)2=n2+n-1.∴a7=72+7-1=55.答案:C4.已知数列{an}为12,13+23,14+24+34,…,110+210+310+…+910,…,如果bn=1anan+1,那么数列{bn}的前n项和Sn为()A.nn+1B.4nn+1C.3nn+1D.5nn+1解析: an=1+2+3+…+nn+1=n2,∴bn=1anan+1=4n(n+1)=4(1n...
