課時(shí)跟蹤檢測(cè)(十四)數(shù)列求和1.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且滿足2a3=a4+3,S7=49.(1)求數(shù)列{an}的通項(xiàng)公式;(2)若數(shù)列{bn}滿足bn=an,n為奇數(shù),2n,n為偶數(shù),求數(shù)列{2.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,且滿足a1+a3+a5=15,S7=49.(1)求{an}的通項(xiàng)公式;(2)若數(shù)列{bn}滿足bn=an·3n,求{bn}的前n項(xiàng)和Tn.3.已知數(shù)列an3n是以13為首項(xiàng)的常數(shù)列,Sn為數(shù)列{an}(1)求Sn;(2)設(shè)正整數(shù)m=b0×30+b1×31+…+bk×3k,其中bi∈{0,1,2},i,k∈N.例如:3=0×30+1×31,則b0=0,b1=1;4=1×30+1×31,則b0=1,b1=1.若f(m)=b0+b1+…+bk,求數(shù)列{Sn·f(Sn)}的前n項(xiàng)和Tn.4.(2024·天津高考)已知數(shù)列{an}是等比數(shù)列,公比大于0,其前n項(xiàng)和為Sn,若a1=1,S2=a31.(1)求數(shù)列{an}的前n項(xiàng)和Sn;(2)設(shè)bn=k,n=ak①當(dāng)k≥2,n=ak+1時(shí),求證:bn1≥ak·bn;②求eq\i\su(i=1,Sn,)bi.課時(shí)跟蹤檢測(cè)(十四)1.解:(1)依題意,設(shè)數(shù)列{an}的公差為d,因?yàn)?a3解得a所以an=a1+(n1)d=1+2(n1)=2n1.(2)因?yàn)閎n=an,n為奇數(shù),2所以T10=b1+b2+…+b9+b10=1+22+5+24+…+17+210=(1+5+…+17)+(22+24+…+210)=5×(1+17)2+22?2121?2.解:(1)設(shè)等差數(shù)列{an}的公差為d,因?yàn)閍1+a3+a5=15,S7=49,所以3a1+6d=15,7a1所以an=1+(n1)×2=2n1.(2)由題意可知bn=(2n1)×3n,所以Tn=1×3+3×32+5×33+…+(2n1)×3n①,3Tn=1×32+3×33+5×34+…+(2n1)×3n+1②,①②得,2Tn=1×31+2×32+2×33+2×34+…+2×3n(2n1)×3n+1=3+2×32?2×3n+11?3(2n1)×3n+1=(2n+2)×3n+16,∴Tn=(3.解:(1)由題意可得an3n=13,則an=3n1,可得an+1an=3n3n?1=3,可知數(shù)列{an}是首項(xiàng)a1=1,公比(2)因?yàn)镾n=30+31+32+…+3n1=1×30+1×31+1×32+…+1×3n1,則b0=b1=…=bn1=1,由題意f(Sn)=b0+b1+…+bn1=1+1+…+1=n,所以Sn·f(Sn)=n×3n?n2,可得Tn=1×3?12+2×32?22+…+n×3n?n2=1(ⅰ)先求數(shù)列{n×3n}的前n項(xiàng)和,記為T',則T'=1×31+2×32+…+n×3n,①3T'=1×32+2×33+…+n×3n+1,②①②得2T'=3+32+33+…+3nn×3n+1=3?3n+1?2n×3n+1=32+所以T'=34+2n?14(ⅱ)再求{n}的前n項(xiàng)和,記為T″,則T″=n(n+1)2.綜上所述,Tn=12(T'T″)=1234+4.解:(1)設(shè){an}的公比為q(q>0),則1+q=q21,解得q=2,所以Sn=1?2n1?2=2(2)①證明:由(1)知,ak=2k1,所以bn=k,n=2k當(dāng)n=ak+1=2k時(shí),bn=k+1,則bn1=b2k?1=b2k?2+2k=b2k?3+4k=…=b2k?1+2k·(2k11)=k+2k·所以bn1ak·bn=k·2kk(k+1)2k1=(k1)2k1k.設(shè)f(x)=(x1)2x1x,x≥2,則f'(x)=2x1+(x1)2x1ln21≥2+2ln21>0,所以f(x)在[2,+∞)上單調(diào)遞增,所以f(x)≥f(2)=0.所以bn1ak·bn≥0,即bn1≥ak·bn.②令k=1,得b1=1,令k=2,得b2=2,b3=b2+2k=6,令k=3,得b4=3,b5=b4+2k=9,b6=b5+2k=15,b7=b6+2k=21,…,所以b2k?1,b2k?1+1,…,b2因?yàn)閎2k?1=k,b2k?1所以b2k?1+b2k?1+1+…+b2所以eq\i\su(i=1,Sn,)bi=eq\i\su(i=1,2n1,)bi=b1+b2+…+b2n?1=1×40+2×41+…+n×4n1,4eq\i\su(i=1,Sn,)bi=1×41+2×42+…+n×4n,兩式相