培優(yōu)課求數(shù)列的通項(xiàng)

培優(yōu)課——求數(shù)列的通項(xiàng)必備知識(shí)基礎(chǔ)練1.將一些數(shù)排成倒三角形如圖所示,其中第一行各數(shù)依次為1,2,3,…,2021,從第二行起,每一個(gè)數(shù)都等于他“肩上”的兩個(gè)數(shù)之和,最后一行只有一個(gè)數(shù)M,則M等于()A.2021×22018 B.2022×22019C.2021×22019 D.2022×220202.(多選題)設(shè)首項(xiàng)為1的數(shù)列{an}的前n項(xiàng)和為Sn,已知Sn+1=2Sn+n1,則下列結(jié)論正確的是()A.數(shù)列{Sn+n}為等比數(shù)列B.數(shù)列{an}的通項(xiàng)公式為an=2n11C.數(shù)列{an+1}為等比數(shù)列D.數(shù)列{SnSn1+1}為等比數(shù)列3.已知在數(shù)列{an}中,a1=1,(2n+1)an=(2n3)an1(n≥2),則數(shù)列{an}的通項(xiàng)公式為.

4.已知Sn是數(shù)列{an}的前n項(xiàng)和,且Sn=2an+n4.(1)求a1的值;(2)若bn=an1,試證明數(shù)列{bn}為等比數(shù)列.5.已知各項(xiàng)均為正數(shù)的數(shù)列{bn}的首項(xiàng)為1,且前n項(xiàng)和Sn滿足SnSn1=Sn+Sn-1(n≥2).試求數(shù)列關(guān)鍵能力提升練6.在數(shù)列{an}中,a1=5,且滿足an+12n-52=an2n-A.2n3 B.2n7C.(2n3)(2n7) D.2n57.已知數(shù)列{an}滿足a1=1,an+1=anan+2(n∈N*).若bn=log21an+1,則數(shù)列{bn}的通項(xiàng)公式bnA.12n B.n1 C.n D.28.若數(shù)列{an}滿足a1=1,且an+1=4an+2n,則a6=.

9.已知在數(shù)列{an}中,an+1=2an+3·2n+1,且a1=2,則數(shù)列{an}的通項(xiàng)公式為.

10.在數(shù)列{an}中,a1=1,a1+2a2+3a3+…+nan=n+12an+1(n∈N*),求數(shù)列{an}的通項(xiàng)公式a11.某企業(yè)投資1000萬(wàn)元用于一個(gè)高科技項(xiàng)目,每年可獲利25%,由于企業(yè)間競(jìng)爭(zhēng)激烈,每年年底需要從利潤(rùn)中取出200萬(wàn)元進(jìn)行廣告投資方能保持原有的利潤(rùn)增長(zhǎng)率.問(wèn)經(jīng)過(guò)多少年后,該項(xiàng)目的資金可以達(dá)到或超過(guò)翻兩番(4倍)的目標(biāo)?(取lg2≈0.30103)學(xué)科素養(yǎng)創(chuàng)新練12.設(shè)關(guān)于x的二次方程anx2an+1x+1=0(n=1,2,3,…)有兩根α和β,且滿足6α2αβ+6β=3.(1)試用an表示an+1;(2)求證:an-(3)當(dāng)a1=76時(shí),求數(shù)列{an}的通項(xiàng)公式

參考答案培優(yōu)課——求數(shù)列的通項(xiàng)1.B記第n行的第一個(gè)數(shù)為an,則a1=1,a2=3=2a1+1,a3=8=2a2+2,a4=20=2a3+4,…,an=2an1+2n2,∴an2n-2=an-12n∴an2n-2=2+(n1)×1=n+1,∴an=(n+1)又每行比上一行的數(shù)字少1個(gè),∴最后一行為第2021行,∴M=a2021=2022×22019.2.AD因?yàn)镾n+1=2Sn+n1,所以Sn+1+又S1+1=2,所以數(shù)列{Sn+n}是首項(xiàng)為2,公比為2的等比數(shù)列,故A正確;所以Sn+n=2n,則Sn=2nn.當(dāng)n≥2時(shí),an=SnSn1=2n11,但a1≠2111,故B錯(cuò)誤;由a1=1,a2=1,a3=3可得a1+1=2,a2+1=2,a3+1=4,即a3+1a2+1由Sn=2nn,所以SnSn1+1=2nn2n1+n1+1=2n1,故D正確.3.an=3(2n-1)(2n+1)由(2n+1)an=(2n3)所以a2a1=15,a3a2=37,a4a3=59,a5又a1=1滿足上式,所以數(shù)列{an}的通項(xiàng)公式為an=3(2n-1)(24.(1)解因?yàn)镾n=2an+n4,所以當(dāng)n=1時(shí),S1=2a1+14,解得a1=3.(2)證明因?yàn)镾n=2an+n4,所以當(dāng)n≥2時(shí),Sn1=2an1+n14,SnSn1=(2an+n4)(2an1+n5),即an=2an11,所以an1=2(an11),又bn=an1,所以bn=2bn1,且b1=a11=2≠0,所以數(shù)列{bn}是以2為首項(xiàng),2為公比的等比數(shù)列.5.解∵SnSn1=Sn+Sn∴(Sn+Sn-1)(Sn-又Sn>0,∴Sn-又S1=∴數(shù)列{Sn}是首項(xiàng)為1,公差為1的等差數(shù)列∴Sn=1+(n1)×1=n,故Sn=n2當(dāng)n≥2時(shí),bn=SnSn1=n2(n1)2=2n1.當(dāng)n=1時(shí),b1=1符合上式.∴bn=2n1.6.C因?yàn)閍n+12n-52=又a12-7=1,所以數(shù)列an2n-7是以1為首項(xiàng),公差為2的等差數(shù)列,所以an2n-7=1+2(n1)7.C由an+1=anan+2,得1a所以1an+1+1=21an又1a1+1=2,所以數(shù)列1an+1是首項(xiàng)為2,公比為2的等比數(shù)列,所以1an+1=2·所以bn=log21an+1=log22n=n.8.2016因?yàn)閍n+1=4an+2n,所以an+1+2n=4(an+2n1),所以數(shù)列{an+2n1}是等比數(shù)列,首項(xiàng)為2,公比為4,則an+2n1=2×4n1,可得an=22n12n1,則a6=22×61261=21125=2016.9.an=(3n2)·2n∵an+1=2an+3·2n+1,∴an+12n+1=a∴數(shù)列an2n是公差為又a12=1,∴an2n=1+3(n1),∴an=(3n10.解由a1+2a2+3a3+…+nan=n+12an+1當(dāng)n≥2時(shí),a1+2a2+3a3+…+(n1)an1=n2an兩式作差得nan=n+12an+1n2得(n+1)an+1=3nan(n≥2),即數(shù)列{nan}從第二項(xiàng)起是公比為3的等比數(shù)列,且a1=1,a2=1,于是2a2=2,故當(dāng)n≥2時(shí),nan=2×3n2.于是an=111.解設(shè)該項(xiàng)目n年后資金數(shù)為an,n∈N*.則由已知得an+1=an(1+25%)200,即an+1=54an200令an+1x=54(anx),即an+1=54an由x4=200,得x=800∴an+1800=54(an800)故數(shù)列{an800}是以a1800為首項(xiàng),54為公比的等比數(shù)列∵a1=1000×(1+25%)200=1050,∴a1800=250,∴an800=250×54n1,∴an=800+250×54n1(n∈N*).由題意知an≥4000,∴800+250×54n1≥4000,即54n≥16兩邊取常用對(duì)數(shù)得nlg54≥lg16,即n(13lg2)≥4lg2∵lg2≈0.30103,∴不等式化為n≥4lg21-3lg2≈∴n≥13.故經(jīng)過(guò)13年后,該項(xiàng)目資金可達(dá)到或超過(guò)翻兩番的目標(biāo).12.(1)解根據(jù)根與系數(shù)的關(guān)系,得α代入題設(shè)條件6(α+β)2αβ=3,得6an+1所以an+1=12an+1(2)證明因?yàn)閍n+1=12an+1所以an+123=12an若an=2