課時(shí)提升作業(yè)(三十)一、選擇題1.已知數(shù)列11×2,12×3,13×4,1n(n+1),下面各數(shù)中是此數(shù)列中的項(xiàng)的是()(A)135(B)142(C)148(D)1542.由a1=1,an+1=an3an+1,給出的數(shù)列an的第34項(xiàng)為()(A)34103(B)100(C)1100(D)11043.(20xx·南昌模擬)已知數(shù)列an的前n項(xiàng)和Sn=2-2n+1,則a3=()(A)-1(B)-2(C)-4(D)-84.已知數(shù)列an的前n項(xiàng)和Sn=2n2-3n+1,則a4+a5+a6+a7+a8+a9+a10的值為()(A)150(B)161(C)160(D)1715.(20xx·西安模擬)在數(shù)列an中,a1=1,anan-1=an-1+(-1)n(n2,nN+),則a3a5的值是()(A)1516(B)158(C)34(D)386.在數(shù)列an中,a1=2,an+1=an+ln(1+1n),則an=()(A)2+lnn(B)2+(n-1)lnn(C)2+nlnn(D)1+n+lnn7.已知數(shù)列an的前n項(xiàng)和Sn=n2-9n,第k項(xiàng)滿足5<ak<8,則k等于()(A)9(B)8(C)7(D)68.(能力挑戰(zhàn)題)定義:F(x,y)=yx(x>0,y>0),已知數(shù)列an滿足:an=F(n,2)F(2,n)(nN+),若對(duì)任意正整數(shù)n,都有anak(kN+)成立,則ak的值為()(A)89(B)2(C)3(D)4二、填空題9.數(shù)列-12,34,-78,1516,的一個(gè)通項(xiàng)公式可以是.10.數(shù)列an的前n項(xiàng)和記為Sn,a1=1,an+1=2Sn+1(n1,nN+),則數(shù)列an的通項(xiàng)公式是.11.(20xx·贛州模擬)已知數(shù)列an滿足a1=12,an-1-an=anan-1n(n-1)(n2),則該數(shù)列的通項(xiàng)公式an=.12.(能力挑戰(zhàn)題)已知數(shù)列an滿足:a1=m(m為正整數(shù)),an+1=an2,當(dāng)an為偶數(shù)時(shí),3an+1,當(dāng)an為奇數(shù)時(shí).若a6=1,則m所有可能的值為.三、解答題13.已知數(shù)列an滿足前n項(xiàng)和Sn=n2+1,數(shù)列bn滿足bn=2an+1,且前n項(xiàng)和為Tn,設(shè)cn=T2n+1-Tn.(1)求數(shù)列bn的通項(xiàng)公式.(2)判斷數(shù)列cn的增減性.14.(能力挑戰(zhàn)題)解答下列各題:(1)在數(shù)列an中,a1=1,an+1=can+cn+1(2n+1)(nN+),其中實(shí)數(shù)c0.求an的通項(xiàng)公式.(2)數(shù)列an滿足:a1=1,an+1=3an+2n+1(nN+),求an的通項(xiàng)公式.15.(20xx·廣東高考)設(shè)數(shù)列an前n項(xiàng)和為Sn,數(shù)列Sn的前n項(xiàng)和為Tn,滿足Tn=2Sn-n2,nN+.(1)求a1的值.(2)求數(shù)列an的通項(xiàng)公式.答案解析1.【解析】選B.42=6×7,故選B.2.【解析】選C.把遞推式取倒數(shù)得1an+1=1an+3,所以1a34=1a1+3×(34-1)=100,所以a34=1100.3.【解析】選D.a3=S3-S2=-14-(-6)=-8.4.【解析】選B.S10-S3=(2×102-3×10+1)-(2×32-3×3+1)=161.5.【解析】選C.當(dāng)n=2時(shí),a2·a1=a1+(-1)2,a2=2.當(dāng)n=3時(shí),a3a2=a2+(-1)3,a3=12.當(dāng)n=4時(shí),a4a3=a3+(-1)4,a4=3.當(dāng)n=5時(shí),a5a4=a4+(-1)5,a5=23,a3a5=34.6.【思路點(diǎn)撥】根據(jù)遞推式采用“疊加”方法求解.【解析】選A.an+1=an+ln(1+1n)=an+lnn+1n=an+ln(n+1)-lnn,a2=a1+ln2,a3=a2+ln3-ln2,an=an-1+lnn-ln(n-1),將上面n-1個(gè)式子左右兩邊分別相加得an=a1+ln2+(ln3-ln2)+(ln4-ln3)+lnn-ln(n-1)=a1+lnn=2+lnn.7.【解析】選B.an=S1,n=1,Sn-Sn-1,n2,即an=-8,n=1,-10+2n,n2.n=1時(shí)也適合an=2n-10,an=2n-10.5<ak<8,5<2k-10<8,152<k<9.又kN+,k=8.8.【解析】選A.an=2nn2,an+1an=2n+1(n+1)22nn2=2n2(n+1)2,2n2-(n+1)2=n2-2n-1,只有當(dāng)n=1,2時(shí),2n2<(n+1)2,當(dāng)n3時(shí),2n2>(n+1)2,即當(dāng)n3時(shí),an+1>an,故數(shù)列an中的最小項(xiàng)是a1,a2,a3中的較小者,a1=2,a2=1,a3=89,故ak的值為89.9.【解析】正負(fù)相間使用(-1)n,觀察可知第n項(xiàng)的分母是2n,分子比分母的值少1,故an=(-1)n2n-12n.答案：an=(-1)n2n-12n10.【思路點(diǎn)撥】根據(jù)an和Sn的關(guān)系轉(zhuǎn)換an+1=2Sn+1(n1)為an+1與an的關(guān)系或者Sn+1與Sn的關(guān)系.【解析】方法一:由an+1=2Sn+1可得an=2Sn-1+1(n2),兩式相減得an+1-an=2an,an+1=3an(n2).又a2=2S1+1=3,a2=3a1,故an是首項(xiàng)為1,公比為3的等比數(shù)列,an=3n-1.方法二:由于an+1=Sn+1-Sn,an+1=2Sn+1,所以Sn+1-Sn=2Sn+1,Sn+1=3Sn+1,把這個(gè)關(guān)系化為Sn+1+12=3(Sn+12),即得數(shù)列Sn+12為首項(xiàng)是S1+12=32,公比是3的等比數(shù)列,故Sn+12=32×3n-1=12×3n,故Sn=12×3n-12.所以,當(dāng)n2時(shí),an=Sn-Sn-1=3n-1,由n=1時(shí)a1=1也適合這個(gè)公式,知所求的數(shù)列an的通項(xiàng)公式是an=3n-1.答案：an=3n-1【方法技巧】an和Sn關(guān)系的應(yīng)用技巧在根據(jù)數(shù)列的通項(xiàng)an與前n項(xiàng)和的關(guān)系求解數(shù)列的通項(xiàng)公式時(shí),要考慮兩個(gè)方面,一個(gè)是根據(jù)Sn+1-Sn=an+1把數(shù)列中的和轉(zhuǎn)化為數(shù)列的通項(xiàng)之間的關(guān)系;一個(gè)是根據(jù)an+1=Sn+1-Sn把數(shù)列中的通項(xiàng)轉(zhuǎn)化為前n項(xiàng)和的關(guān)系,先求Sn再求an.11.【解析】由遞推公式變形,得1an-1an-1=1n(n-1)=1n-1-1n,則1a2-1a1=1-12,1a3-1a2=12-13,1an-1an-1=1n-1-1n,各式相加得1an-1a1=1-1n,即1an=3n-1n,an=n3n-1.答案：n3n-112.【解析】根據(jù)遞推式以及a1=m(m為正整數(shù))可知數(shù)列an中的項(xiàng)都是正整數(shù).a6=1,若a6=a52,則a5=2,若a6=3a5+1,則a5=0,故只能是a5=2.若a5=a42,則a4=4,若a5=3a4+1,則a4=13,故只能是a4=4.若a4=a32,則a3=8,若a4=3a3+1,則a3=1.(1)當(dāng)a3=8時(shí),若a3=a22,則a2=16,若a3=3a2+1,則a2=73,故只能是a2=16,若a2=a12,則a1=32,若a2=3a1+1,則a1=5.(2)當(dāng)a3=1時(shí),若a3=a22,則a2=2,若a3=3a2+1,則a2=0,故只能是a2=2.若a2=a12,則a1=4,若a2=3a1+1,則a1=13,故只能是a1=4.綜上所述:a1的值,即m的值只能是4或5或32.答案：4或5或32【變式備選】已知數(shù)列an中,a1=12,an+1=1-1an(n2),則a16=.【解析】由題可知a2=1-1a1=-1,a3=1-1a2=2,a4=1-1a3=12,此數(shù)列為循環(huán)數(shù)列,a1=a4=a7=a10=a13=a16=12.答案：1213.【解析】(1)a1=2,an=Sn-Sn-1=2n-1(n2).bn=1n,n2,nN+,23,n=1.(2)cn=bn+1+bn+2+b2n+1=1n+1+1n+2+12n+1,cn+1-cn=12n+2+12n+3-1n+1=-n-1(2n+2)(2n+3)(n+1)<0,cn是遞減數(shù)列.14.【解析】(1)由原式得an+1cn+1=ancn+(2n+1).令bn=ancn,則b1=1c,bn+1=bn+(2n+1),因此對(duì)n2有bn=(bn-bn-1)+(bn-1-bn-2)+(b2-b1)+b1=(2n-1)+(2n-3)+3+1c=n2-1+1c,因此an=(n2-1)cn+cn-1,n2.又當(dāng)n=1時(shí)上式成立.因此an=(n2-1)cn+cn-1,nN+.(2)兩端同除以2n+1得,an+12n+1=32·an2n+1,即an+12n+1+2=32(an2n+2),即數(shù)列an2n+2是首項(xiàng)為a121+2=52,公比為32的等比數(shù)列,故an2n+2=52×(32)n-1,即an=5×3n-1-2n+1.15.【解析】(1)當(dāng)n=1時(shí),T1=2S1-1.因?yàn)門1=S1=a1,所以a1=2a1-1,求得a1=1.(2)當(dāng)n2時(shí),Sn=Tn-Tn-1=2Sn-n2-2Sn-1-(n-1)2=2Sn-2Sn-1-2n+1,所以Sn=2Sn-1+2n-1,所以Sn+1=2Sn+2n+1,-得an+1=2an+2,所以an+1+2=2(an+2),即an+1+2an+2=2(n2),求得a1+2=3,a2+2=6,則a2+2a1+2=2.所以an+2是以3為首項(xiàng),2為公比的等比數(shù)列,所以an+2=3·2n-1,所以an=3·2n-1-2,nN+.