【金版学案】2015-2016学年高中数学2.3.1数列前n项和与等差数列的前n项和练习新人教A版必修5►基础梳理1.(1)对于任意数列{an},Sn=__________________,叫做数列{an}的前n项的和.(2)Sn-Sn-1=____________.2.(1)等差数列{an}的前n项和公式为________________________________________________________________________.(2)等差数列:2,4,6,…,2n,…的前n项和Sn=__________.(3)等差数列首项为a1=3,公差d=-2,则它的前6项和为______.3.(1)等差数列依次k项之和仍然是等差数列.即Sk,S2k-Sk,S3k-S2k,…成公差为______________的等差数列.(2)已知等差数列{an},an=n,则S3,S6-S3,S9-S6分别为:________.它们成______数列.4.(1)由Sn的定义可知,当n=1时,S1=________;当n≥2时,an=__________,即an=__________________.(2)已知等差数列{an}的前n项和为Sn=n2,则an=________________=____________.5.(1)等差数列的前n项和公式:Sn=na1+可化成关于n的二次式子为________________________,当d≠0时,是一个常数项为零的二次式.(2)已知等差数列的前n项和为Sn=n2-8n,则前n项和的最小值为______,此时n=______.基础梳理1.(1)a1+a2+a3+…+an(2)an(n≥2),a1=S1(n=1)2.(1)Sn=或Sn=na1+(2)(n+1)n(3)-123.(1)k2d(2)6,15,24等差4.(1)a1Sn-Sn-1(2)2n-1,n∈N*5.(1)Sn=n2+n(2)-164►自测自评1.(2014·福建卷)等差数列{an}的前n项和Sn,若a1=2,S3=12,则a6=()A.8B.10C.12D.142.已知数列{an}中,a+a+2a3a8=9,且an<0,则S10为()A.-9B.-11C.-13D.-153.1+4+7+10+…+(3n+4)+(3n+7)等于()A.B.C.D.自测自评1.解析:设公差为d,依题意得3×2+×3×2d=12,∴d=2,所以a6=2+(6-1)×2=112,故选C.答案:C2.解析:(a3+a8)2=9, an<0,∴a3+a8=-3.∴S10==-15.答案:D3.解析:本题的项数为n+3项,这一点很关键.答案:C►基础达标1.已知a1,a2,a3,a4成等差数列,若S4=32,a2∶a3=1∶3,则公差d为()A.8B.16C.4D.01.解析:S4=32⇒2(a2+a3)=32,∴a2+a3=16,又=,a3=3a2,∴a2=4,a3=12,∴d=a3-a2=8.故选A.答案:A2.设a1,a2,…和b1,b2,…都是等差数列,其中a1=25,b1=75,a100+b100=100,则数列{an+bn}前100项之和为()A.0B.100C.10000D.505002.解析:S100=×100=10000.故选C.答案:C3.等差数列{an}中,首项a1>0,公差d<0,Sn为其前n项和,则点(n,Sn)可能在下列哪条曲线上()3.解析:由Sn=na1+n(n-1)d=n2+n,及d<0,a1>0知,<0,a1->0,故排除A,B.对称轴n=-=>0,排除D.答案:C4.已知等差数列共有2n+1项,其中奇数项之和为290,偶数项之和为261,则an+1的值为()A.30B.29C.28D.274.解析:奇数项共有n+1项,其和为×(n+1)=·(n+1)=290,∴(n+1)an+1=290,偶数项共有n项,其和为×n=·n=nan+1=261,∴an+1=290-261=29.故选B.答案:B5.(2013·上海卷)若等差数列的前6项和为23,前9项和为57,则数列的前n项和Sn=________.5.n2-n►巩固提高26.已知两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,且=,则的值为()A.B.C.D.6.解析:S2n-1=(2n-1)·=(2n-1)·=(2n-1)an.同理T2n-1=(2n-1)bn.∴==.令n=11得===.故选C.答案:C7.已知lgx+lgx3+lgx5+…+lgx21=11,则x=________________________________________________________________________.7.解析:由条件得lg(x·x3·x5·…·x21)=11⇒lgx1+3+5+…+21=11⇒121lgx=11,lgx=,x=10.答案:8.已知数列{an}的前n项和Sn=4n2+2(n∈N*),则an=______________________.8.解析:n=1时,a1=S1=6;n≥2时,an=Sn-Sn-1=4n2-4(n-1)2=8n-4.∴an=答案:9.在小于100的正整数中共有多个数被3除余2?这些数的和是多少?9.分析:被3除余2的正整数可以写成3n+2(n∈N*)的形式.解析:由3n+2<100,得n<32,即n=0,1,2,3,…,32.∴在小于100的正整数中共有33个数被3除余2.把这些数从小到大排列起来为:2,5...