52.课标全国卷数列试题的求和“情结”之求和递推.doc

杨培明(手机号码:13965261699)数学丛书,给您一个智慧的人生！课标高考全国卷数学试题揭秘.预测 第6讲:课标全国卷数列试题的求和“情结”之求和递推 229 第6讲:课标全国卷数列试题的求和“情结”之求和递推特色惊爆 课标全国卷中数列试题的求和“情结”的一种重要表现形式是研究an与Sn的关系递推(即求和递推)式;an与Sn的关系:当n=1时,a1=S1;当n2时,an=Sn-Sn-1;该关系式是解决an与Sn的关系递推式问题的基础工具.情结渊源1.(2014年课标高考试题理科第17题)已知数列an的前n项和为Sn,a1=1,an0,anan+1=Sn-1,其中为常数.()证明:an+2-an=; ()是否存在,使得an为等差数列？并说明理由.解析:()由anan+1=Sn-1an+1an+2=Sn+1-1,两式相减得:an+1(an+2-an)=an+1(an+10)an+2-an=;()由a1=1,anan+1=Sn-1a2=-1;又由an+2-an=a2n-1与a2n均是公差为的等差数列a2n-1=n-+1,a2n=n-1an=-+1=n+(n为奇数),an=n-1(n为偶数);所以,an为等差数列=-1=4,故存在=4,使得an为等差数列.2.(2015年课标高考试题理科第17题)Sn为数列an的前n项和.已知an>0,an2+2an=4Sn+3.()求an的通项公式; ()设bn=,求数列bn的前n项和.解析:()由an2+2an=4Sn+3a12+2a1=4S1+3a1=S1=3(-1舍去);又由an2+2an=4Sn+3an+12+2an+1=4Sn+1+3,两式相减得:(an+12+2an+1)-(an2+2an)=4an+1(an+1+an)(an+1-an-2)=0an+1-an=2数列an是公差d=2的等差数列an=2n+1;()由an=2n+1bn=(-)数列bn的前n项和Tn=(-)+(-)+(-)=.命题规律 数列an的前n项和为Sn,解决an与Sn的关系递推式问题的方法有:相减消元:由Sn=f(an)an=Sn-Sn-1=f(an)-f(an-1),得关于an的递推关系;代入消元:由F(an,Sn)=0F(Sn-Sn-1,Sn)=0,得关于Sn的递推关系;母题结论有:如果Sn=Aan+B(n1)其中A0,1,B0,则an是等比数列;如果a1=a且Sn=Aan(n2),其中aA0,A1,则an(n2)是等比数列,但an不是等比数列;若an+an+10且2dSn=an2+dan-a(a+d)(n1),则an是等差数列,且an=dn+a或an=dn-d-a.原创预测 1.相减消元: 230 第6讲:课标全国卷数列试题的求和“情结”之求和递推 原创示例:设Sn为数列an的前n项和,已知a10,2an-a1=S1Sn,nN+.()求数列an的通项公式;()设bn=,求数列bn的前n项和Tn.解析:()由2an-a1=S1Sna1=S12a1=a12a1=1;当n2时,(2an-1)-(2an-1-1)=Sn-Sn-1=anan=2an-1an=2n-1;()由an=2n-1Sn=2n-1bn=-Tn=b1+b2+bn=(-)+(-)+(-)+(-)=-=1-=.原创预测:1.已知数列an的前n项和为Sn,a1=1,an+1=2Sn+1(n1).()求an的通项公式()等差数列bn的各项为正,其前n项和为Tn,且T3=15,又a1+b1,a2+b2,a3+b3成等比数列,求Tn.2.已知数列an的前n项和为Sn,且满足a1=a(a0),an+1=rSn(nN+,rR,r-1).()求数列an的通项公式;()若存在kN+,使得Sk+1,Sk,Sk+2成等差数列,试判断:对任意的mN+,且m2,am+1,am,am+2是否成等差数列,并证明你的结论.3.已知各项均为正数的数列an的前n项和Sn满足:S1>1且6Sn=(an+1)(an+2)(n1).()求an的通项公式; ()令bn=,求数列bn的前n项和Tn.4.已知正项数列an,其前n项和sn满足:10Sn=an2+5an+6,且a1,a3,a15成等比数列.()求数列an的通项an; ()求最小的正整m,使得a1,a3,a15,am(m>15)成等比数列.5.设各项均为正数的数列an的前n项和为Sn,已知a1=1,=an+1-n2-n-,nN+.()求数列an的通项公式; ()证明:对一切正整数n,有+<. 2.代入消元:原创示例:已知数列an的前n项和为Sn满足:S1=1,当n2时,2an=anSn-Sn2.()证明:数列成等差数列; ()证明:数列Sn2的前n项和Tn<4.解析:()当n2时,由2an=anSn-Sn22(Sn-Sn-1)=(Sn-Sn-1)Sn-Sn22(Sn-Sn-1)=-Sn-1Sn成等差数列;()由Sn=Sn2=<=4(-)Tn<4(1-)<4.原创预测:6.已知数列an的前n项和为Sn满足:S1=1,当n2时,an=.()求数列an的通项公式; ()若数列的前n项和为Tn,问满足Tn>的最小正整数n是多少？7.设Sn是数列an的前n项和,且a1=-1,an+1=SnSn+1.()求数列an的通项公式; ()证明:数列Sn2的前n项和Tn<28.设数列an的前n项和为Sn,且方程x2-anx-an=0有一根为Sn-1,n=1,2,3,.()求an的通项公式;()证明:当n2时,an=(Sn-1)(Sn-1-1). 第6讲:课标全国卷数列试题的求和“情结”之求和递推 231 9.设数列an的前n项和为Sn,数列Sn的前n项和为Tn,满足Tn=Sn-n(n+1),nN+.()求数列an的通项公式; ()证明:+<.10.已知正数数列an满足a1=1,Sn=(an+),其中,Sn为数列an的前n项和.()求数列an的通项公式; ()对于实数x,x表示不超过实数x的最大整数,求+.原创解析:1.解:()由an+1=2Sn+1an+1-an=(2Sn+1)-(2Sn-1+1)=2anan+1=3anan=3n-1;()由T3=3b2=15b2=5;又由(a1+b1)(a3+b3)=(a2+b2)2(1+b1)(9+b3)=64(6-d)(14+d)=64d=2(-10舍去)b1=3Tn=n2+2n.2.解:()由an+1=rSna2=rS1=ra1=ra;当n2时,an+1-an=r(Sn-Sn-1)=ranan+1=(1+r)anan=a2(1+r)n-2=ar(1+r)n-2;而a1=a不适合该式,所以,an=;()由an+1=rSnar(1+r)n-1=rSnSn=a(1+r)n-1;所以,Sk+1,Sk,Sk+2成等差数列Sk+1+Sk+2=2Ska(1+r)k+a(1+r)k+1=2a(1+r)k-1(1+r)+(1+r)2=2r=0,或r=-3;当r=0,且m2时,am=0am+1,am,am+2成等差数列;当r=-3,且m2时,am=-3a(-2)m-2am+1+am+2=-3a(-2)m-1-3a(-2)m=-3a(-2)m-1+(-2)m=-6a(-2)m-2=2amam+1,am,am+2成等差数列.3.解:()由6Sn=(an+1)(an+2)6S1=(a1+1)(a2+2)6a1=(a1+1)(a2+2)a1=2;当n2时,6an=6(Sn-Sn-1)=(an+1)(an+2)-(an-1+1)(an-1+2)(an+an-1)(an-an-1-3)=0an-an-1=3an=3n-1;()由an=3n-1bn=-Tn=-=.4.解:()由10Sn=an2+5an+610S1=a12+5a1+6a1=S1=2或3;当n2时,10an=10Sn-10Sn-1=(an2+5an+6)-(an-12+5an-1+6)(an+an-1)(an-an-1-5)=0an-an-1=5an=5n-3或an=5n-2;当an=5n-3时,a1=2,a3=12,a15=72成等比数列;当an=5n-2时,a1=3,a3=13,a15=73不成等比数列,故an=5n-3;()由等比数列2,12,72,的通项=2×6n-1am=5m-3=2×6n-15m=2×6n-1+3,当n=1,2,3时,m=1,3,15,分别得到a1,a3,a15,当n=4时,m=87最小的正整m=87.5.解:()当n2时,2an=2Sn-2Sn-1=n(an+1-n2-n-)-(n-1)an-(n-1)2-(n-1)-nan+1=(n+1)an+n(n+1)=+1=nan=n2;()由=<=-+<1+(-)+(-)+(-)=1+(-)<.6.解:()当n2时,由an=Sn-Sn-1=1=nSn=n2an=2n-1;()由=(-)Tn=(1-)=;又由Tn>>n>最小正整数n=504.7.解:由a1=-1S1=-1;又由an+1=SnSn+1Sn+1-Sn=SnSn+1-=-1是公差d=-1的等差数列=-nSn=-;当n2时,an=Sn-Sn-1=-,a1=-1不适合该式. 232 第6讲:课标全国卷数列试题的求和“情结”之求和递推 ()由Sn=-Sn2=;当n2时,Sn2=<=-Tn<1+(1-)<2.8.解:()由方程x2-anx-an=0有一根为Sn-1(Sn-1)2-an(Sn-1)-an=0S1=,(Sn-1)2-(Sn-Sn-1)(Sn-1)-(Sn-Sn-1)=0SnSn-1-2Sn+1=0Sn=-1=-n-1Sn=an=;()由Sn=1-1-Sn=1-Sn-1=an=(1-Sn)(1-Sn-1)=(Sn-1)(Sn-1-1).9.解:()由Tn=Sn-n(n+1)T1=S1=1;当n2时,Sn=Tn-Tn-1=Sn-n(n+1)-Sn-1-n(n-1)Sn=3Sn-1+nSn+=3(Sn-1+)Sn+=×3n-1Sn=×3n-1-an=;()由an=,作公比为q的等比数列xn,使得:x1=1,且=q=xn=()n-1;因xn()n-123n-13n-13n-11成立,所以,+数列xn的前n项和=1-()n<.10.解:()由Sn=(an+)2Sn=(Sn-Sn-1)+Sn2-Sn-12=1Sn=;当n2时,an=Sn-Sn-1=-;()由Sn=;又由<<2(-)<<2(-)2(-1)<+<1+2(10-1)+=18.预测例证