题型练4大题专项二高中数学数列的通项求和问题

1题型练4大题专项(二)数列的通项、求和问题题型练第64页一、解答题1.已知数列{an}满足a2-a1=1,其前n项和为Sn,当n≥2时,Sn-1-1,Sn,Sn+1成等差数列.(1)求证{an}为等差数列;(2)若Sn=0,Sn+1=4,求n.答案:(1)证明当n≥2时,由Sn-1-1,Sn,Sn+1成等差数列,可知2Sn=Sn-1-1+Sn+1,即Sn-Sn-1=-1+Sn+1-Sn,即an=-1+an+1(n≥2),则an+1-an=1(n≥2),又a2-a1=1,故{an}是公差为1的等差数列.(2)解由(1)知等差数列{an}的公差为1.由Sn=0,Sn+1=4,得an+1=4,即a1+n=4.由Sn=0,得na1+-=0,即a1+-=0,解得n=7.2.已知等差数列{an}满足a4=7,2a3+a5=19.(1)求an;(2)设{bn-an}是首项为2,公比为2的等比数列,求数列{bn}的通项公式及其前n项和Tn.解:(1)由题意得{解得{∴an=1+2(n-1)=2n-1.(2)由题意可知bn-an=2n,∴bn=2n+2n-1,∴Tn=(2+22+…+2n)+[1+3+…+(2n-1)],∴Tn=2n+1+n2-2.3.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+n.(1)求q的值;(2)求数列{bn}的通项公式.解:(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8()=20,解得q=2或q=,因为q1,所以q=2.(2)设cn=(bn+1-bn)an,数列{cn}的前n项和为Sn,2由cn={--解得cn=4n-1.由(1)可知an=2n-1,所以bn+1-bn=(4n-1)()-故bn-bn-1=(4n-5)()-,n≥2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+…+(b3-b2)+(b2-b1)=(4n-5)()-+(4n-9)()-+…+7+3.设Tn=3+7+11()+…+(4n-5)()-,n≥2,Tn=3+7()+…+(4n-9)()-+(4n-5)()-,所以Tn=3+4+4()+…+4()--(4n-5)()-,因此Tn=14-(4n+3)()-,n≥2,又b1=1,所以bn=15-(4n+3)()-4.已知等差数列{an}的前n项和为Sn,公比为q的等比数列{bn}的首项是,且a1+2q=3,a2+4b2=6,S5=40.(1)求数列{an},{bn}的通项公式an,bn;(2)求数列{}的前n项和Tn.解:(1)设{an}公差为d,由题意得{解得{故an=3n-1,bn=()(2)(-)(-)+22n+1,∴Tn=[(-)(-)+…+(--)]--(-)(22n+3-8)=(-)35.已知数列{an}满足a1=,且an+1=an-(n∈N*).(1)证明:12(n∈N*);(2)设数列{}的前n项和为Sn,证明:(n∈N*).答案：证明(1)由题意得an+1-an=-0,即an+1≤an,故an由an=(1-an-1)an-1,得an=(1-an-1)(1-an-2)…(1-a1)a10.由0an,得--[1,2],即12.(2)由题意得=an-an+1,所以Sn=a1-an+1.①由和12,得12,所以n2n,因此an+1(n∈N*).②由①②得(n∈N*).6.已知等比数列{an}的前n项和为Sn,且an+1=1+Sn,且a2=2a1.(1)求数列{an}的通项公式;(2)若bn=anlog2an+(-1)n·n,求数列{bn}的前n项和Hn.解：(1)∵an+1=1+Sn,∴当n≥2时,an=1+Sn-1,∴an+1=2an(n≥2).又a2=1+S1=1+a1,a2=2a1,解得a1=1.∴an=2n-1.(2)由题意可知bn=anlog2an+(-1)n·n=(n-1)·2n-1+(-1)n·n.设数列{(n-1)·2n-1}的前n项和为Tn,则有Tn=0×20+1×21+2×22+…+(n-1)·2n-1,①∴2Tn=0×21+1×22+2×23+…+(n-1)·2n,②由②-①,得Tn=(n-2)·2n+2.当n为偶数时,Hn=(n-2)·2n+2-1+2-3+…-(n-1)+n=(n-2)·2n+2+=(n-2)·2n+4当n为奇数时,Hn=(n-2)·2n+2-1+2-3+…-(n-1)-n=(n-2)·2n+2+--n=(n-2)·2n--故Hn={-为偶数---为奇数