【志鴻優(yōu)化設(shè)計(jì)】高考數(shù)學(xué)一輪復(fù)習(xí) 第六章 數(shù)列考點(diǎn)規(guī)范練28 數(shù)列的概念與表示文.doc

考點(diǎn)規(guī)范練28數(shù)列的概念與表示一、非標(biāo)準(zhǔn)1.數(shù)列0,的一個(gè)通項(xiàng)公式為()a.an=(nn+)b.an=(nn+)c.an=(nn+)d.an=(nn+)2.若sn為數(shù)列an的前n項(xiàng)和,且sn=,則等于()a.b.c.d.303.設(shè)數(shù)列an滿足:a1=2,an+1=1-,記數(shù)列an的前n項(xiàng)之積為tn,則t2015的值為()a.-b.-1c.d.24.已知數(shù)列an的前n項(xiàng)和為sn,a1=1,sn=2an+1,則sn等于()a.2n-1b.c.d.5.數(shù)列an滿足:a1+3a2+5a3+(2n-1)an=(n-1)3n+1+3(nn+),則數(shù)列an的通項(xiàng)公式an=.6.已知數(shù)列an的通項(xiàng)公式為an=(n+2),則當(dāng)an取得最大值時(shí),n=.7.設(shè)an是首項(xiàng)為1的正項(xiàng)數(shù)列,且(n+1)-n+an+1an=0(n=1,2,3,),則它的通項(xiàng)公式an=.8.設(shè)數(shù)列an的前n項(xiàng)和為sn.已知a1=1,=an+1-n2-n-,nn+.(1)求a2的值;(2)求數(shù)列an的通項(xiàng)公式.9.已知函數(shù)f(x)=ax2+bx(a0)的導(dǎo)函數(shù)f(x)=-2x+7,數(shù)列an的前n項(xiàng)和為sn,點(diǎn)pn(n,sn)(nn+)均在函數(shù)y=f(x)的圖象上,求數(shù)列an的通項(xiàng)公式及sn的最大值.10.(2014湖南長沙模擬)已知函數(shù)f(x)是定義在(0,+)上的單調(diào)函數(shù),且對(duì)任意的正數(shù)x,y都有f(xy)=f(x)+f(y).若數(shù)列an的前n項(xiàng)和為sn,且滿足f(sn+2)-f(an)=f(3)(nn+),則an等于()a.2n-1b.nc.2n-1d.11.已知數(shù)列an滿足:a1=1,an+1=(nn+).若bn+1=(n-),b1=-,且數(shù)列bn是單調(diào)遞增數(shù)列,則實(shí)數(shù)的取值范圍為()a.2b.3c.2d.0,(n+1)an+1-nan=0,即,=,an=.8.解:(1)依題意,2s1=a2-1-,又s1=a1=1,所以a2=4.(2)由題意2sn=nan+1-n3-n2-n,所以當(dāng)n2時(shí),2sn-1=(n-1)an-(n-1)3-(n-1)2-(n-1),兩式相減得,2an=nan+1-(n-1)an-(3n2-3n+1)-(2n-1)-,整理得(n+1)an-nan+1=-n(n+1),即=1.又=1,故數(shù)列是首項(xiàng)為=1,公差為1的等差數(shù)列,所以=1+(n-1)1=n,所以an=n2.9.解:f(x)=ax2+bx(a0),f(x)=2ax+b.又f(x)=-2x+7,a=-1,b=7.f(x)=-x2+7x.點(diǎn)pn(n,sn)(nn+)均在函數(shù)y=f(x)的圖象上,sn=-n2+7n.當(dāng)n=1時(shí),a1=s1=6;當(dāng)n2時(shí),an=sn-sn-1=-2n+8,a1=6適合上式,an=-2n+8(nn+).令an=-2n+80得n4,當(dāng)n=3或n=4時(shí),sn取得最大值12.綜上,an=-2n+8(nn+),且當(dāng)n=3或n=4時(shí),sn取得最大值12.10.d解析:由題意知f(sn+2)=f(an)+f(3)=f(3an)(nn+),sn+2=3an,sn-1+2=3an-1(n2),兩式相減得,2an=3an-1(n2).又n=1時(shí),s1+2=3a1=a1+2,a1=1.數(shù)列an是首項(xiàng)為1,公比為的等比數(shù)列.an=.11. c解析:由已知可得+1,+1=2.又+1=20,則+1=2n,bn+1=2n(n-),bn=2n-1(n-1-)(n2).b1=-也適合上式,故bn=2n-1(n-1-)(nn+).由bn+1bn,得2n(n-)2n-1(n-1-),即n+1恒成立.而n+1的最小值為2,故的取值范圍為2.12.n2-n解析:觀察圖象,發(fā)現(xiàn)a1=1,a2=a1+4,a3=a2+7,a4=a3+10,猜測(cè)當(dāng)n2時(shí),an=an-1+3n-2,an-an-1=3n-2.an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=(3n-2)+(32-2)+1=n2-n.13.解:(1)an=a1=,an=.(2)當(dāng)n4時(shí),6,an=,當(dāng)n5時(shí),10,an=.從第5項(xiàng)開始及其以后各項(xiàng)均小于.14.解:(1)當(dāng)n=1時(shí),t1=2s1-1.t1=s1=a1,a1=2a1-1.a1=1.(2)當(dāng)n2時(shí),tn-1=2sn-1-(n-1)2,則sn=tn-tn-1=2sn-n2-=2(sn-sn-1)-2n+1=2an-2n+1.因?yàn)楫?dāng)n=1時(shí),a1=s1=1也滿足上式,所以sn=2an-2n+1(n1).當(dāng)n2時(shí),sn-1=2an-1-2(n-1)+1,兩式相減得an=2an-2an-1-2,所以an=2an-1+2(n2)