1、加课练(4)数列求和1.已知等差数列an和等比数列bn满足a1=b1=1,a2+a4=10,b2b4=a5.(1)求an的通项公式;(2)求和:b1+b3+b5+b2n-1.解析(1)设等差数列an的公差为d.因为a1=1,a2+a4=10,所以2a1+4d=10,解得d=2.所以an=2n-1.(2)设等比数列bn的公比为q.因为b2b4=a5,所以b1qb1q3=9.又b1=1,所以q2=3.所以b2n-1=b1q2n-2=3n-1.从而b1+b3+b5+b2n-1=1+3+32+3n-1=3n-12.2.Sn为数列an的前n项和.已知an0,an2+2an=4Sn+3.(1)求an的通项
2、公式;(2)设bn=1anan+1,求数列bn的前n项和.解析(1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.-,得an+12-an2+2(an+1-an)=4an+1,即2(an+1+an)=an+12-an2=(an+1+an)(an+1-an).由an0,得an+1-an=2.又a12+2a1=4a1+3,所以a1=-1(舍去)或a1=3.所以an是首项为3,公差为2的等差数列,通项公式为an=2n+1.(2)由(1)知an=2n+1,则bn=1anan+1=1(2n+1)(2n+3)=1212n+1-12n+3.设数列bn的前n项和为Tn,则Tn=b1+
3、b2+bn=1213-15+15-17+12n+1-12n+3=1213-12n+3=n3(2n+3).3.(2018天津文,18,13分)设an是等差数列,其前n项和为Sn(nN*);bn是等比数列,公比大于0,其前n项和为Tn(nN*),已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;(2)若Sn+(T1+T2+Tn)=an+4bn,求正整数n的值.解析(1)设等比数列bn的公比为q(q0).由b1=1,b3=b2+2,可得q2-q-2=0.由q0,可得q=2,故bn=2n-1.所以Tn=1-2n1-2=2n-1.设等差数列an的公差为d.由b4=a
4、3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,从而a1=1,d=1,故an=n,所以Sn=n(n+1)2.(2)由(1),有T1+T2+Tn=(21+22+2n)-n=2(1-2n)1-2-n=2n+1-n-2.由Sn+(T1+T2+Tn)=an+4bn可得n(n+1)2+2n+1-n-2=n+2n+1,整理得n2-3n-4=0,解得n=-1(舍去)或n=4.所以,n的值为4.4.(2018浙江,20,15分)已知等比数列an的公比q1,且a3+a4+a5=28,a4+2是a3,a5的等差中项.数列bn满足b1=1,数列(bn+1-bn)an的前n项和为2n2+
5、n.(1)求q的值;(2)求数列bn的通项公式.解析(1)由a4+2是a3,a5的等差中项得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20得8q+1q=20,解得q=2或q=12,因为q1,所以q=2.(2)设cn=(bn+1-bn)an,数列cn的前n项和为Sn.由cn=S1,n=1,Sn-Sn-1,n2,解得cn=4n-1.由(1)可知an=2n-1,所以bn+1-bn=(4n-1)12n-1,故bn-bn-1=(4n-5)12n-2,n2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+(b3-b2)+(b2-b1)=(4n-5)12n-2+(4n-9)12n-3+712+3.设Tn=3+712+11122+(4n-5)12n-2,n2,12Tn=312+7122+(4n-9)12n-2+(4n-5)12n-1,所以12Tn=3+412+4122+412n-2-(4n-5)12n-1,因此Tn=14-(4n+3)12n-2,n2,又b1=1,所以bn=15-(4n+3)12n-2.
