20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：高考大題專項(三)-數(shù)列-【含解析】(編輯修改稿)
2025-04-03 02:52 本頁面
【文章內(nèi)容簡介】 (1)數(shù)列不等式的證明要把數(shù)列的求和與放縮法結(jié)合起來,即放縮程度越小,保留的項就越少,運算就越簡單.(2)證明數(shù)列不等式也經(jīng)常轉(zhuǎn)化為數(shù)列和的最值問題,同時要注意比較法、放縮法、基本不等式的應(yīng)用.高考大題專項(三) 數(shù)列例1解(1)設(shè)數(shù)列{an}的公差為d,由a2=3,a4+a5+a6=0,得a2=3,a5=0,即a1+d=3,a1+4d=0,解得a1=4,d=1.數(shù)列{an}的通項公式an=a1+(n1)d=5n,前n項和Sn=na1+n(n1)2d=4n+n(n1)2(1)=9nn22.(2)由an=5n,得a1=4,a2=3,a3=2,a4=1.由題意知應(yīng)取b1=a1=4,b2=a3=2,b3=a4=1.所以數(shù)列{bn}的公比q=12,Tn=b1(1qn)1q=4[1(12)n]112=∈N*,所以4≤Tn8.又由(1)知Sm=9mm22(m∈N*).由此知,當(dāng)m=4,5時,Sm取得最大值10,要使SmTn+k恒成立,只需使104+k即可,所以有k6,由k是正整數(shù)知,k的最小值為7.對點訓(xùn)練1解(1)∵an是Sn與2的等差中項,∴Sn=2an2,①∴Sn1=2an12(n≥2,n∈N*),②由①②,得SnSn1=2an2an1(n≥2,n∈N*),∴an=2an2an1,∵an≠0,∴anan1=2(n≥2,n∈N*),即數(shù)列{an}是等比數(shù)列.∵Sn=2an2,∴a1=S1=2a12,解得a1=2.∴an=2n.∵點P(bn,bn+1)在直線xy+2=0上,∴bnbn+1+2=0.∴bn+1bn=2,即{bn}=1,∴bn=2n1.(2)∵=2n+2n1,∴Tn=(2+22+23+…+2n)+(1+3+5+…+2n1),∴Tn=2(12n)12+n(1+2n1)2=2n+1+n22.例2解(1)當(dāng)n1時,Sn1=32an11.∴an=SnSn1=32an132an11.∴an=3an1,∵a1=2≠0,∴數(shù)列{an}是等比數(shù)列,∴an=23n1,∴Sn=3n1.(2)數(shù)列Sn+1Sn是遞減數(shù)列.證明:設(shè)bn=Sn+1Sn,∴bn=3n+113n1=3+23n1.∴bn+1bn=23n+1123n1=2(3n3n+1)(3n+11)(3n1).∵n≥1,∴3n+13n1,∴3n+13n0,3n+110,3n10,∴bn+1bn0.∴{bn}是遞減數(shù)列,即數(shù)列Sn+1Sn是遞減數(shù)列.對點訓(xùn)練2(1)解當(dāng)n=1時,a1=1。當(dāng)n≥2時,a1+2a2+3a3+…+nan=14[(2n1)3n+1],a1+2a2+3a3+…+(n1)an1=14[(2n3)3n1+1].兩式相減,得an=3n1,當(dāng)n=1時,a1也符合上式,綜上可得an=3n1(n∈N*).(2)證明由(1)可知bn=123n11≤13n1(n∈N*),故b1+b2+…+bn130+131+132+…+13n1=32113n3
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