1、课时作业31数列求和一、选择题1(2014武汉质检)已知数列an的通项公式是an,其前n项和Sn,则项数n()A13B10C9 D6解析:an1,Snnn1,n6.答案:D2(2014西安质检)已知数列an满足a11,an1an2n(nN*),则S2 012()A22 0121 B321 0063C321 0061 D321 0052解析:a11,a22,又2.2.a1,a3,a5,成等比数列;a2,a4,a6,成等比数列,S2 012a1a2a3a4a5a6a2 011a2 012(a1a3a5a2 011)(a2a4a6a2 012)321 0063.故选B.答案:B3(2014杭州模拟)
2、已知函数f(x)x22bx过(1,2)点,若数列的前n项和为Sn,则S2 012的值为()A. B.C. D.解析:由已知得b,f(n)n2n,S2 01211.答案:D4(2014西安模拟)数列an满足anan1(nN*),且a11,Sn是数列an的前n项和,则S21()A. B6C10 D11解析:依题意得anan1an1an2,则an2an,即数列an中的奇数项、偶数项分别相等,则a21a11,S21(a1a2)(a3a4)(a19a20)a2110(a1a2)a211016,故选B.答案:B5(2014长沙模拟)已知函数f(n)n2cos(n),且anf(n)f(n1),则a1a2a3
3、a100()A100 B0C100 D10 200解析:若n为偶数时,则anf(n)f(n1)n2(n1)2(2n1),为首项为a25,公差为4的等差数列;若n为奇数,则anf(n)f(n1)n2(n1)22n1,为首项为a13,公差为4的等差数列所以a1a2a3a100(a1a3a99)(a2a4a100)503450(5)4100.答案:A6(2014广东广州综合测试一)在数列an中,已知a11,an1ansin,记Sn为数列an的前n项和,则S2 014()A1 006 B1 007C1 008 D1 009解析:由an1ansinan1ansin,所以a2a1sin101,a3a2si
4、n1(1)0,a4a3sin2000,a5a4sin011,因此a5a1,如此继续可得an4an(nN*),数列an是一个以4为周期的周期数列,而2 01445032,因此S2 014503(a1a2a3a4)a1a2503(1100)111 008,故选C.答案:C二、填空题7(2015山西晋中名校联考)在数列an中,a11,an1(1)n(an1),记Sn为an的前n项和,则S2 013_.解析:由a11,an1(1)n(an1)可得a11,a22,a31,a40,该数列是周期为4的数列,所以S2 013503(a1a2a3a4)a2 013503(2)11 005.答案:1 0058(2
5、014武汉模拟)等比数列an的前n项和Sn2n1,则aaa_.解析:当n1时,a1S11,当n2时,anSnSn12n1(2n11)2n1,又a11适合上式an2n1,a4n1.数列a是以a1为首项,以4为公比的等比数列aaa(4n1)答案:(4n1)9(2014广东揭阳一模)对于每一个正整数n,设曲线yxn1在点(1,1)处的切线与x轴的交点的横坐标为xn,令anlgxn,则a1a2a99_.解析:曲线yxn1在点(1,1)处的切线方程为y(n1)(x1)1,即y(n1)xn,它与x轴交于点(xn,0),则有(n1)xnn0xn,anlgxnlglgnlg(n1),a1a2a99(lg1lg
6、2)(lg2lg3)(lg99lg100)lg1lg1002.答案:2三、解答题10(2014新课标全国卷)已知an是递增的等差数列,a2,a4是方程x25x60的根(1)求an的通项公式;(2)求数列的前n项和解析:(1)方程x25x60的两根为2,3,由题意得a22,a43.设数列an的公差为d,则a4a22d,故d,从而a1.所以an的通项公式为ann1.(2)设的前n项和为Sn,由(1)知,则Sn,Sn.两式相减得Sn.所以Sn2.11(2014安徽卷)数列an满足a11,nan1(n1)ann(n1),nN*.(1)证明:数列是等差数列;(2)设bn3n,求数列bn的前n项和Sn.解
7、析:(1)由已知可得1,即1.所以是以1为首项,1为公差的等差数列(2)由(1)得1(n1)1n,所以ann2.从而bnn3n.Sn131232333n3n,3Sn132233(n1)3nn3n1.得2Sn31323nn3n1n3n1.所以Sn.12(2014湖南卷)已知数列an的前n项和Sn,nN*.(1)求数列an的通项公式;(2)设bn2an(1)nan,求数列bn的前2n项和解析:(1)当n1时,a1S11;当n2时,anSnSn1n.故数列an的通项公式为ann.(2)由(1)知,bn2n(1)nn.记数列bn的前2n项和为T2n,则T2n(212222n)(12342n)记A212222n,B12342n,则A22n12,B(12)(34)(2n1)2nn.故数列bn的前2n项和T2nAB22n1n2.