高考數(shù)學(xué)理科人教通用版核心講練大一輪復(fù)習(xí)練考題預(yù)測(cè)全過(guò)關(guān)5-4數(shù)列求和

溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)鼠標(biāo)滾軸，調(diào)節(jié)合適的觀看比例，答案解析附后。關(guān)閉Word文檔返回原板塊。練考題預(yù)測(cè)·全過(guò)關(guān)1.(2019·天津高考)設(shè){an}是等差數(shù)列,{bn}是等比數(shù)列,公比大于0,已知a1=b1=3,b2=a3,b3=4a2+3.(1)求{an}和{bn}的通項(xiàng)公式.(2)設(shè)數(shù)列{cn}滿足cn=1,n為奇數(shù),bn2,n為偶數(shù),求a1c1+a2c2【解題指南】(1)首先設(shè)出等差數(shù)列的公差,等比數(shù)列的公比,根據(jù)題意,列出方程組,求出公差和公比,進(jìn)而求得等差數(shù)列和等比數(shù)列的通項(xiàng)公式.(2)根據(jù)題中所給的cn所滿足的條件,將a1c1+a2c2+…+a2nc2n表示出來(lái),之后應(yīng)用分組求和法,結(jié)合等差數(shù)列的求和公式,以及錯(cuò)位相減法求和,最后求得結(jié)果.【解析】(1)設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q,依題意,得3q=3+2故an=3+3(n1)=3n,bn=3×3n1=3n,所以{an}的通項(xiàng)公式為an=3n,{bn}的通項(xiàng)公式為bn=3n.(2)a1c1+a2c2+…+a2nc2n=(a1+a3+a5+…+a2n1)+(a2b1+a4b2+a6b3+…+a2nbn)=n×3+n(n-1)2×6=3n2+6×(1×31+2×32+…+n×3n),記Tn=1×31+2×32+…+n×3n①則3Tn=1×32+2×33+…+n×3n+1②②①得2Tn=33233…3n+n×3n+1=3(1-3n所以a1c1+a2c2+…+a2nc2n=3n2+6Tn=3n2+3×(=(2n-1)2.已知數(shù)列{an}滿足:當(dāng)n≥2且n∈N*時(shí),有an+an1=(1)n×3.則數(shù)列{an}的前200項(xiàng)的和為 ()【解析】選A.當(dāng)n≥2且n∈N*時(shí),有an+an1=(1)n×3.可得a2+a1=3,a4+a3=3,a6+a5=3,…,a200+a199=3,則數(shù)列{an}的前200項(xiàng)的和為:(a1+a2)+(a3+a4)+…+(a199+a200)=3×100=300.3.已知Sn為數(shù)列{an}的前n項(xiàng)和,an=2·3n1(n∈N*),若bn=an+1SnSn+1,則b1+b2【解析】由an=2·3n1可知,數(shù)列{an}是以2為首項(xiàng),3為公比的等比數(shù)列,所以Sn=2(1-則bn=an+1SnSn則b1+b2+…+bn=1S1-1S2+1S2-1S答案:124.設(shè)Sn=11×2+12×3+13×4+…+1n(n+1)(n∈N【解析】Sn=11×2+12×3+13×4+…+1n(n+1)=112+由SnSn+1=56,可得nn+1·n+1n解得n=10.答案:105.數(shù)列{an}的通項(xiàng)公式為an=92(n=1),n2·3n________.

【解析】數(shù)列{an}的前n項(xiàng)和Sn=92+22·32+32·33+42·34+…+n2·3n3Sn=272+22·33+32·34+42·35+…+n2·3n+1相減可得2Sn=27+5·33+7·34+…+(2n1)·3nn2·3n+1,設(shè)Tn=5·33+7·34+…+(2n1)·3n,3Tn=5·34+7·35+…+(2n1)·3n+1,相減可得2Tn=5·33+2(34+…+3n)(2n1)·3n+1=135+2·81(1-3n化簡(jiǎn)可得Tn=(