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2019.5
專(zhuān)題升級(jí)訓(xùn)練 解答題專(zhuān)項(xiàng)訓(xùn)練(數(shù)列)
1.設(shè)數(shù)列{an}的前n項(xiàng)和Sn滿(mǎn)足2Sn=an+1-2n+1+1,n∈N*,且a1,a2+5,a3成等差數(shù)列.
(1)求a1的值;
(2)求數(shù)列{an}的通項(xiàng)公式.
2.已知各項(xiàng)都不相等的等差數(shù)列{an}的前6項(xiàng)和為60,且a6為a1和a21的等比中項(xiàng).
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)若數(shù)列{bn}滿(mǎn)足bn+1-bn=an(n∈N*),且b1=3,求數(shù)列的前n項(xiàng)和Tn.
3.已知數(shù)列{an}是公差為正的等差數(shù)列,其前n項(xiàng)和為
2、Sn,點(diǎn)(n,Sn)在拋物線(xiàn)y=x2+x上;各項(xiàng)都為正數(shù)的等比數(shù)列{bn}滿(mǎn)足b1b3=,b5=.
(1)求數(shù)列{an},{bn}的通項(xiàng)公式;
(2)記Cn=anbn,求數(shù)列{Cn}的前n項(xiàng)和Tn.
4.已知Sn是等比數(shù)列{an}的前n項(xiàng)和,S4,S10,S7成等差數(shù)列.
(1)求證a3,a9,a6成等差數(shù)列;
(2)若a1=1,求數(shù)列{}的前n項(xiàng)的積.
5.已知數(shù)列{an}滿(mǎn)足:a1=1,an+1=
(1)求a2,a3;
(2)設(shè)bn=a2n-2,n∈N*,求證:{bn}是等比數(shù)列,并求其通項(xiàng)公式;
(3)在(2)的條件下,求數(shù)列{an}前100項(xiàng)中的所有偶數(shù)項(xiàng)的和S.
3、6.已知數(shù)列{an}(n∈N*)是首項(xiàng)為a,公比為q≠0的等比數(shù)列,Sn是數(shù)列{an}的前n項(xiàng)和,已知12S3,S6,S12-S6成等比數(shù)列.
(1)當(dāng)公比q取何值時(shí),使得a1,2a7,3a4成等差數(shù)列?
(2)在(1)的條件下,求Tn=a1+2a4+3a7+…+na3n-2.
7.已知數(shù)列{an}的各項(xiàng)排成如圖所示的三角形數(shù)陣,數(shù)陣中每一行的第一個(gè)數(shù)a1,a2,a4,a7,…構(gòu)成等差數(shù)列{bn},Sn是{bn}的前n項(xiàng)和,且b1=a1=1,S5=15.
(1)若數(shù)陣中從第三行開(kāi)始每行中的數(shù)按從左到右的順序均構(gòu)成公比為正數(shù)的等比數(shù)列,且公比相等,已知a9=16,求a50的值;
4、(2)設(shè)Tn=+…+,求Tn.
8.(20xx廣東深圳模擬,18)各項(xiàng)為正數(shù)的數(shù)列{an}滿(mǎn)足=4Sn-2an-1(n∈N*),其中Sn為{an}的前n項(xiàng)和.
(1)求a1,a2的值;
(2)求數(shù)列{an}的通項(xiàng)公式;
(3)是否存在正整數(shù)m,n,使得向量a=(2an+2,m)與向量b=(-an+5,3+an)垂直?說(shuō)明理由.
##
1.解:(1)在2Sn=an+1-2n+1+1中,
令n=1,得2S1=a2-22+1,
令n=2,得2S2=a3-23+1,[來(lái)源:]
解得a2=2a1+3,a3=6a1+13.
又2(a2+5)=a1+a3,解得a1=1.
(2)2Sn=
5、an+1-2n+1+1,2Sn+1=an+2-2n+2+1,得an+2=3an+1+2n+1,
又a1=1,a2=5也滿(mǎn)足a2=3a1+21,
∴an+1=3an+2n對(duì)n∈N*成立.
∴an+1+2n+1=3(an+2n),
∴an+2n=3n,∴an=3n-2n.
2.解:(1)設(shè)等差數(shù)列{an}的公差為d(d≠0),
則解得∴an=2n+3.
(2)由bn+1-bn=an,
∴bn-bn-1=an-1(n≥2,n∈N*),
bn=(bn-bn-1)+(bn-1-bn-2)+…+(b2-b1)+b1
=an-1+an-2+…+a1+b1=(n-1)+3=n(n+2).
6、
∴bn=n(n+2)(n∈N*).
∴,
Tn=
=
=.
3.解:(1)∵Sn=n2+n,
當(dāng)n=1時(shí),a1=S1=2;
當(dāng)n≥2時(shí),Sn-1=(n-1)2+(n-1)=n2-n+1.
∴an=Sn-Sn-1=3n-1(n≥2).
當(dāng)n=1時(shí),a1=3-1=2滿(mǎn)足題意.
∴數(shù)列{an}是首項(xiàng)為2,公差為3的等差數(shù)列.
∴an=3n-1.
又∵各項(xiàng)都為正數(shù)的等比數(shù)列{bn}滿(mǎn)足b1b3=,b5=,
∴b2=b1q=,b1q4=,
解得b1=,q=,∴bn=.
(2)∵Cn=(3n-1),
∴Tn=2+5+…+(3n-4)+(3n-1),①
∴Tn=2+5+…
7、+(3n-4)+(3n-1),②
①-②,得Tn=1+3-(3n-1)
=1+3-(3n-1)-3-(3n-1).
∴Tn=5-.
4.解:(1)當(dāng)q=1時(shí),2S10≠S4+S7,∴q≠1.
由2S10=S4+S7,得.
∵a1≠0,q≠1,∴2q10=q4+q7.[來(lái)源:]
則2a1q8=a1q2+a1q5.∴2a9=a3+a6.
∴a3,a9,a6成等差數(shù)列.
(2)依題意設(shè)數(shù)列{}的前n項(xiàng)的積為T(mén)n,
Tn=
=13q3(q2)3…(qn-1)3=q3(q3)2…(q3)n-1
=(q3)1+2+3+…+(n-1)=(q3.
又由(1)得2q10=q4+q7,
8、
∴2q6-q3-1=0,解得q3=1(舍),q3=-.
∴Tn=.
5.解:(1)a2=,a3=-.
(2)
=,
又b1=a2-2=-,
∴數(shù)列{bn}是等比數(shù)列,且bn==-.
(3)由(2)得a2n=bn+2=2-(n=1,2,3,…,50),
S=a2+a4+…+a100=250-=100-1+=99+.
6.解:(1)由題意可知,a≠0.
①當(dāng)q=1時(shí),則12S3=36a,S6=6a,S12-S6=6a,
此時(shí)不滿(mǎn)足條件12S3,S6,S12-S6成等比數(shù)列;
②當(dāng)q≠1時(shí),則
12S3=12,S6=,S12-S6=,
由題意得12,
化簡(jiǎn)整理得(4q
9、3+1)(3q3-1)(1-q3)(1-q6)=0,
解得q3=-,或q3=,或q=-1.
當(dāng)q=-1時(shí),a1+3a4=-2a,2a7=2a,
∴a1+3a4≠2(2a7),不滿(mǎn)足條件;
當(dāng)q3=-時(shí),a1+3a4=a(1+3q3)=,2(2a7)=4aq6=,
即a1+3a4=2(2a7),∴當(dāng)q=-時(shí),滿(mǎn)足條件;
當(dāng)q3=時(shí),a1+3a4=a(1+3q3)=2a,2(2a7)=4aq6=,
∴a1+3a4≠2(2a7),從而當(dāng)q3=時(shí),不滿(mǎn)足條件.
綜上,當(dāng)q=-時(shí),使得a1,2a7,3a4成等差數(shù)列.
(2)由(1)得na3n-2=na.∴Tn=a+2a+3a+…+(n
10、-1)a+na,①
則-Tn=a+2a+3a+…+(n-1)a+na,②
①-②得Tn=a+a+a+a+…+a-na=a-a,
∴Tn=a-a.
7.解:(1)∵{bn}為等差數(shù)列,設(shè)公差為d,b1=1,S5=15,[來(lái)源:]
∴S5=5+10d=15,d=1.
∴bn=1+(n-1)1=n.
設(shè)從第3行起,每行的公比都是q,且q>0,a9=b4q2,4q2=16,q=2,1+2+3+…+9=45,故a50是數(shù)陣中第10行第5個(gè)數(shù),
而a50=b10q4=1024=160.
(2)∵Sn=1+2+…+n=,
∴Tn=+…+
=+…+
=2
=2.
8.解:(1)
11、當(dāng)n=1時(shí),=4S1-2a1-1,
即(a1-1)2=0,解得a1=1,
當(dāng)n=2時(shí),=4S2-2a2-1=4a1+2a2-1=3+2a2,
解得a2=3或a2=-1(舍去).
(2)由已知=4Sn-2an-1,①
=4Sn+1-2an+1-1,②
②-①得=4an+1-2an+1+2an=2(an+1+an),[來(lái)源:]
即(an+1-an)(an+1+an)=2(an+1+an).
∵數(shù)列{an}各項(xiàng)均為正數(shù),
∴an+1+an>0,
∴an+1-an=2.
∴數(shù)列{an}是首項(xiàng)為1,公差為2的等差數(shù)列,∴an=2n-1.
(3)∵an=2n-1,
∴a=(2an+2,m)=(2(2n+3),m)≠0,b=(-an+5,3+an)=(-(2n+9),2(n+1))≠0.
又a⊥b?ab=0
?m(n+1)-(2n+3)(2n+9)=0
?m=4(n+1)+16+,
∵m,n∈N*,
∴n+1=7,m=47+16+1,即n=6,m=45.[來(lái)源:]
∴當(dāng)且僅當(dāng)n=6,m=45時(shí),a⊥b.
解答題專(zhuān)項(xiàng)訓(xùn)練(數(shù)列)
