专题突破练15求数列的通项及前n项和1.已知等差数列{an}的前n项和为Sn,且a1=1,S3+S4=S5.(1)求数列{an}的通项公式;(2)令bn=(-1)n-1an,求数列{bn}的前n项和Tn.2.(2020山东滨州二模,18)已知{an}为等差数列,a3+a6=25,a8=23,{bn}为等比数列,且a1=2b1,b2b5=a11.(1)求{an},{bn}的通项公式;(2)记cn=an·bn,求数列{cn}的前n项和Tn.3.(2020全国Ⅲ,理17)设数列{an}满足a1=3,an+1=3an-4n.(1)计算a2,a3,猜想{an}的通项公式并加以证明;(2)求数列{2nan}的前n项和Sn.4.(2020山东聊城二模,17)已知数列{an}的各项均为正数,其前n项和为Sn,且an2+an=2Sn+34(n∈N*).(1)求数列{an}的通项公式;(2)若bn=1Sn,求{bn}的前n项和Tn.5.(2020山东青岛5月模拟,17)设数列{an}的前n项和为Sn,a1=1,,给出下列三个条件;条件①:数列{an}为等比数列,数列{Sn+a1}也为等比数列;条件②:点(Sn,an+1)在直线y=x+1上;条件③:2na1+2n-1a2+…+2an=nan+1.试在上面的三个条件中任选一个,补充在上面的横线上,完成下列两问的解答:(1)求{an}的通项公式;(2)设bn=1log2an+1·log2an+3,求数列{bn}的前n项和Tn.6.(2020山东菏泽一模,18)已知数列{an}满足nan+1-(n+1)an=1(n∈N*),且a1=1.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=an3n-1,求数列{bn}的前n项和Sn.7.已知数列{an}的前n项和Sn=3n2+8n,{bn}是等差数列,且an=bn+bn+1.(1)求数列{bn}的通项公式;(2)令cn=(an+1)n+1(bn+2)n,求数列{cn}的前n项和Tn.8.(2020天津南开区一模,18)已知数列{an}的前n项和Sn=n2+n2,数列{bn}满足:b1=b2=2,bn+1bn=2n+1(n∈N*).(1)求数列{an},{bn}的通项公式;(2)求∑i=1nai(b2i-1-1b2i)(n∈N*).专题突破练15求数列的通项及前n项和1.解(1)设等差数列{an}的公差为d,由S3+S4=S5可得a1+a2+a3=a5,即3a2=a5,∴3(1+d)=1+4d,解得d=2.∴an=1+(n-1)×2=2n-1.(2)由(1)可得bn=(-1)n-1·(2n-1).∴T2n=1-3+5-7+…+(2n-3)-(2n-1)=(-2)×n=-2n.∴当n为偶数时,Tn=-n;当n为奇数时,Tn=Tn-1+bn=-(n-1)+(-1)n-1an=-(n-1)+(-1)n-1(2n-1)=-(n-1)+(2n-1)=n.综上,Tn=(-1)n+1n.2.解(1)设等差数列{an}的公差为d,由题意得{2a1+7d=25,a1+7d=23,解得{a1=2,d=3.所以数列{an}的通项公式an=3n-1.设等比数列{bn}的公比为q,由a1=2b1,b2b5=a11,得b1=1,b12q5=32,解得q=2,所以数列{bn}的通项公式bn=2n-1.(2)由(1)知,cn=anbn=(3n-1)×2n-1,则Tn=c1+c2+c3+…+cn-1+cn=2×20+5×21+8×22+…+(3n-4)×2n-2+(3n-1)×2n-1,2Tn=2×21+5×22+8×23+…+(3n-4)×2n-1+(3n-1)×2n.两式相减得-Tn=2+3(21+22+…+2n-1)-(3n-1)×2n=2+3×2-2n-1×21-2-(3n-1)×2n=-4+(4-3n)×2n,所以Tn=4+(3n-4)×2n.3.解(1)a2=5,a3=7.猜想an=2n+1.由已知可得an+1-(2n+3)=3[an-(2n+1)],an-(2n+1)=3[an-1-(2n-1)],……a2-5=3(a1-3).因为a1=3,所以an=2n+1.(2)由(1)得2nan=(2n+1)2n,所以Sn=3×2+5×22+7×23+…+(2n+1)×2n.①从而2Sn=3×22+5×23+7×24+…+(2n+1)×2n+1.②①-②得-Sn=3×2+2×22+2×23+…+2×2n-(2n+1)×2n+1.所以Sn=(2n-1)2n+1+2.4.解(1)因为an2+an=2Sn+34(n∈N*),①所以当n≥2时,an-12+an-1=2Sn-1+34,②①-②得an2−an-12+an-an-1=2(Sn-Sn-1),即an2−an-12-an-an-1=0,所以(an+an-1)(an-an-1-1)=0.因为an>0,所以an-an-1=1,所以数列{an}是公差为1的等差数列,当n=1时,由a12+a1=2S1+34可得,a1=32,所以an=a1+(n-1)d=32+(n-1)×1=n+12.(2)由(1)知Sn=na1+n(n-1)2d=n(n+2)2,所以bn=1Sn=2n(n+2)=1n−1n+2,Tn=b1+b2+b3+…+bn-1+bn=(11-13)+(12-14)+13−15+…+(1n-1-1n+1)+(1n-1n+2)=32−1n+1−1n+2=3n2+5n2n2+6n+4.5.解(1)方案一:选条件①.因为数列{Sn+a1}为等比数列,所以(S2+a1)2=(S1+a1)(S3+a1),即(2a1+a2)2=2a1(2a1+a2+a3).设等比数列{an}的公比为q,因为a1=1,所以(2+q)2=2(2+q+q2),解得q=2或q=0(舍).所以an=a1qn-1=2n-1(n∈N*).方案二:选条件②.因为点(Sn,an+1)在直线y=x+1上,所以an+1=Sn+1(n∈N*),所以an=Sn-1+1(n≥2).两式相减得an+1-an=an,an+1an=2(n≥2).因为a1=1,a2=S1+1=a1+1=2,a2a1=2适合上式,所以数列{an}是首项为1,公比为2的等比数列,所以an=a1qn-1=2n-1(n∈N*).方案三:选条件③.当n≥2时,因为2na1+2n-1a2+…+2an=nan+1(...
