適用于新高考新教材2024版高考數(shù)學二輪復習考點突破練5數(shù)列求和及其綜合應用課件

考點突破練5數(shù)列求和及其綜合應用1234561.(2023安徽蕪湖高三統(tǒng)考)已知Sn是數(shù)列{an}的前n項和,2Sn=(n+1)an,且a1=1.(1)求{an}的通項公式;(2)設a0=0,已知數(shù)列{bn}滿足

,求{bn}的前n項和Tn.123456

所以Tn=(tan

1-tan

0)+(tan

2-tan

1)+(tan

3-tan

2)+…+[tan

n-tan(n-1)]=tan

n-tan

0=tan

n.1234562.(2023山東濟南二模)已知數(shù)列{an}的前n項和Sn=2n+1-2,數(shù)列{bn}滿足bn=log2an.(1)求數(shù)列{an},{bn}的通項公式;(2)由an,bn構(gòu)成的n×n階數(shù)陣如圖所示,求該數(shù)陣中所有項的和Tn.123456解

(1)因為Sn=2n+1-2,當n=1時,S1=22-2=2,即a1=2,當n≥2時,Sn-1=2n-2,所以Sn-Sn-1=(2n+1-2)-(2n-2),即an=2n,經(jīng)檢驗,當n=1時,an=2n也成立,所以an=2n,則bn=log2an=log22n=n.(2)由數(shù)陣可知Tn=a1(b1+b2+…+bn)+a2(b1+b2+…+bn)+…+an(b1+b2+…+bn)=(a1+a2+…+an)·(b1+b2+…+bn),1234563.(2023河北張家口高三期末)已知Sn為數(shù)列{an}的前n項和,Sn=2an-4n+2.(1)證明:數(shù)列{an+4}為等比數(shù)列;(2)求數(shù)列{nan}的前n項和Tn.123456(1)證明

由題知Sn=2an-4n+2,所以a1=S1=2a1-4×1+2,解得a1=2,故a1+4=6.由Sn=2an-4n+2,可得Sn-1=2an-1-4(n-1)+2,n≥2,兩式相減得an=Sn-Sn-1=2an-2an-1-4,n≥2,所以數(shù)列{an+4}是以6為首項,2為公比的等比數(shù)列.123456(2)解

由(1)得數(shù)列{an+4}是以6為首項,2為公比的等比數(shù)列,所以an+4=6×2n-1,故an=3×2n-4,則nan=3n×2n-4n.設bn=n×2n,其前n項和為Pn,則Pn=1×2+2×22+3×23+…+n×2n,①2Pn=1×22+2×23+3×24+…+n×2n+1,②1234564.(2023廣東河源高三期末)已知等差數(shù)列{an}的前n項和為Sn,a5=5,S8=36,an=log3bn.(1)求{an}和{bn}的通項公式;1234561234561234565.(2023新高考Ⅰ,20)設等差數(shù)列{an}的公差為d,且d>1.令

,記Sn,Tn分別為數(shù)列{an},{bn}的前n項和.(1)若3a2=3a1+a3,S3+T3=21,求{an}的通項公式;(2)若{bn}為等差數(shù)列,且S99-T99=99,求d.1234561234561234561234566.(2023山東青島高三期末)記數(shù)列{an}的前n項和為Sn,a1=1,__________.給出下列兩個條件:條件?,數(shù)列{an}和數(shù)列{Sn+a1}均為等比數(shù)列;條件?,2na1+2n-1a2+…+2an=nan+1.試在上面的兩個條件中任選一個,補充在上面的橫線上,完成下列兩問的解答.

(1)求數(shù)列{an}的通項公式;(2)記正項數(shù)列{bn}的前n項和為Tn,b1=a2,b2=a3,4Tn=bn·bn+1,求(注:如果選擇多個條件分別解答,按第一個解答計分)123456解

(1)若選條件?:∵數(shù)列{Sn+a1}為等比數(shù)列,∴(S2+a1)2=(S1+a1)(S3+a1),即(2a1+a2)2=2a1(2a1+a2+a3).∵a1=1,且設等比數(shù)列{an}的公比為q,∴(2+q)2=2(2+q+q2),解得q=2或q=0(舍去),∴an=a1qn-1=2n-1.若選條件?:∵2na1+2n-1a2+…+2an=nan+1,①∴2n-1a1+2n-2a2+…+2an-1=(n-1)an(n≥2),∴2na1+2n-1a2+…+22an-1=2(n-1)an(n≥2),②①-②得2an=nan+1-2(n-1)an(n≥2),即an+1=2an(n≥2),令2na1+2n-1a2+…+2an=nan+1中n=1,得a2=2a1也符合上式,故數(shù)列{an}為首項a1=1,公比q=2的等比數(shù)列,則an=a1qn-1=2n-1.123456(2)由(1)可知,不論條件為?還是?,都有數(shù)列{an}是首項a1=1,公比q=2的等比數(shù)列,即an=2n-1,則b1=a2=2,b2=a3=4,∵4Tn=bn·bn+1,③∴4Tn-1=bn-1·bn(n≥2),④③-④得4(Tn-Tn-1)=bn·bn+1-bn-1·bn(n≥2),即4bn=bn·(bn+1-bn-1)(n≥2),∵數(shù)列{bn}為正項數(shù)列,則bn+1-bn-1=4(n≥2),則數(shù)列{bn}的奇數(shù)項、偶數(shù)項分別都是公差為4的等差數(shù)列,∴{b2k