專題突破練7數(shù)列求和的方法1.(13分)(2025山東濱州模擬)已知數(shù)列{an}的前n項和Sn=n2+2n.(1)求數(shù)列{an}的通項公式;(2)求數(shù)列2an+1Sn2.(13分)(2025江蘇揚州模擬)已知數(shù)列{an}中,a1=1,Sn為數(shù)列{an}的前n項和,滿足Sn+1=3Sn+n+1(n∈N*).(1)證明:數(shù)列an+12是等比數(shù)列,并求數(shù)列{a(2)設bn=log3(2an+1),求數(shù)列(2an+1)(23.(15分)(2025廣東深圳模擬)已知數(shù)列{an}的前n項和為Sn,a1=t(t≠-1),an+1-Sn=n.(1)當t為何值時,數(shù)列{an+1}是等比數(shù)列?(2)設數(shù)列{bn}的前n項和為Tn,b1=1,點(Tn+1,Tn)在直線xn+1-yn=12上,在(1)的條件下,若不等式b1a1+1+b2a2+14.(15分)(2025浙江臺州模擬)對于數(shù)列{an},記區(qū)間(1,an)內(nèi)偶數(shù)的個數(shù)為bn,則稱數(shù)列{bn}為{an}的偶數(shù)列.(1)若數(shù)列{dn}為數(shù)列{n3}的偶數(shù)列,求d3.(2)若數(shù)列{cn}為數(shù)列{2n+1+3}的偶數(shù)列,證明:數(shù)列{cn-1}為等比數(shù)列.(3)在(2)的前提下,若數(shù)列{bn}為等差數(shù)列{an}的偶數(shù)列,a1=5,a5=13,求數(shù)列{bncn}的前n項和Sn.

答案：1.解(1)因為Sn=n2+2n,所以當n=1時,a1=S1=3.當n≥2時,an=Sn-Sn-1=n2+2n-(n-1)2-2(n-1)=2n+1.當n=1時,上式也成立.所以an=2n+1.(2)由(1)得,2an+1Sn=22n+1+12(1n-1n+2),所以Tn=23+12(1-13)+25+12(12-14)+…+22n+1+12(1n-1n+2),所以Tn=(23+25+…+22n+1)+12.解(1)由題意,當n=1時,S2=3S1+2,得a1+a2=3a1+2.∵a1=1,∴a2=4.當n≥2時,Sn+1=3Sn+n+1,①Sn=3Sn-1+n,②①-②得an+1=3an+1(n≥2).∵a2=4=3a1+1,∴an+1=3an+1(n≥1).則an+1+12=3an+32=3(an+12),∵a1+12∴an+12是以a1+∴an+12=3n2,(2)由(1)得bn=log3(2an+1)=log33n=n,則(2∴(2an+1)(2n-1)bnbn+13.解(1)由an+1-Sn=n,得an-Sn-1=n-1(n≥2),兩式相減得an+1-an-(Sn-Sn-1)=1,即an+1=2an+1,所以an+1+1=2(an+1)(n≥2),由a1=t及an+1-Sn=n,得a2=t+1.因為數(shù)列{an+1}是等比數(shù)列,所以只需要a2+1a1+1=t+2t+1=2,解得t=0,此時,數(shù)列{an+1}是以(2)由(1)得an=2n-1-1,因為點(Tn+1,Tn)在直線xn+1-yn=12上,所以Tn+1n+1-Tnn=12,故Tnn是以T當n≥2時,bn=Tn-Tn-1=n(nb1=1滿足該式,所以bn=n(n∈N*).不等式b1a1+1+b2即為1+22+322+…令Rn=1+22+322則12Rn=12+22兩式相減得12Rn=1+12+122+12所以Rn=4-n+2由Rn≥m-92n恒成立,即4-2n又(4-2n-32n+1)-(4故當n≤3時,4-2當n≥4時,4-2當n=3時,4-2×3-523=318;當n=4時,4-2×4-4.(1)解在區(qū)間(1,33)內(nèi)的偶數(shù)為2,4,6,8,10,12,14,16,18,20,22,24,26,共13個,所以d3=13.(2)證明在區(qū)間(1,2n+1+3)內(nèi)的偶數(shù)為2,4,…,2n+1,2n+1+2,則cn=2n+1+2-22+1=2n+1.于是c1-1=2,cn+1-1cn-1(3)解依題意,等差數(shù)列{an}的公差d=a5-a15-1=2,則an=5+2(n-1)=2n+3,b由(2)知,cn=2n+1,則bncn=(n+1)(2n+1)=(n+1)·2n+(n+1),令數(shù)列{(n+1)·2n}的前n項和為Tn,則Tn=2×2+3×22+…+(n+1)·2n,于是2Tn=2×22+3×23+…+n·2n+(n+1)·2n+1,兩式相減得-Tn=4