2024-2025學(xué)年高中數(shù)學(xué)第二章數(shù)列2.5等比數(shù)列的前n項和第二課時數(shù)列求和習(xí)題課課時作業(yè)新人教A版必修5

PAGE1-其次課時數(shù)列求和(習(xí)題課)[選題明細(xì)表]學(xué)問點(diǎn)、方法題號公式法、并項轉(zhuǎn)化法求和1,4,10分組轉(zhuǎn)化法求和7,8,12裂項相消法求和2,3,6,9錯位相減法求和5,11基礎(chǔ)鞏固1.等比數(shù)列{an}中,a2=9,a5=243,則{an}的前4項和為(B)(A)81 (B)120 (C)168 (D)192解析:因為a5=a2q3,所以q3=a5a2所以q=3,所以a1=3,所以S4=3(2.數(shù)列{an}的通項公式是an=1n(A)11 (B)99 (C)120 (D)121解析:因為an=1n+n+1=所以Sn=a1+a2+…+an=(2-1)+(3-2)+…+(n+1-n)=n令n+1得n=120.故選C.3.已知數(shù)列an=14n2-1(n∈(A)2021 (B)1819 (C)10解析:an=14n2-1=1(2n所以S10=12(11-13+13-15+…+1194.在數(shù)列{an}中,已知Sn=1-5+9-13+17-21+…+(-1)n-1(4n-3),則S15+S22-S31的值為(B)(A)13 (B)-76 (C)46 (D)76解析:因為S15=(-4)×7+(-1)14(4×15-3)=29.S22=(-4)×11=-44.S31=(-4)×15+(-1)30(4×31-3)=61.所以S15+S22-S31=29-44-61=-76.故選B.5.(2024·九江高二月考)數(shù)列{n·2n}的前n項和等于(B)(A)n·2n-2n+2 (B)n·2n+1-2n+1+2(C)n·2n+1-2n (D)n·2n+1-2n+1解析:設(shè){n·2n}的前n項和為Sn,則Sn=1×21+2×22+3×23+…+n·2n,①所以2Sn=1×22+2×23+…+(n-1)·2n+n·2n+1,②①-②得-Sn=2+22+23+…+2n-n·2n+1=2(1-2n所以Sn=n·2n+1-2n+1+2,故選B.6.已知數(shù)列{an}是通項an和公差都不為零的等差數(shù)列,設(shè)Sn=1a1a2+1a2a3+…+1an(A)na1(C)n-1解析:因為{an}是等差數(shù)列,所以1anan+1=1d所以Sn=1d(1a1-1a2+1a2-1a3+…+1an-1an7.(2024·德州高二檢測)求和:Sn=1+(1+12)+(1+12+14)+(1+12+14+18)+…+(1+12+14+解析:被求和式的第k項為ak=1+12+14+…+12k-1=1所以Sn=2[(1-12)+(1-122)+…+(=2[n-(12+122+123=2[n-12=2]n-(1-12=2n+12答案:2n+128.(2024·陜西咸陽期末)已知數(shù)列{an}是等差數(shù)列,{bn}是等比數(shù)列,且b2=3,b3=9,a1=b1,a14=b4.(1)求{an}的通項公式;(2)設(shè)cn=an+bn,求數(shù)列{cn}的前n項和Tn.解:(1)設(shè)數(shù)列{an}的公差為d,{bn}的公比為q,由b2=3,b3=9,得q=b3b2=3,bn=b2qn-2=3·3n-2即有a1=b1=1,a14=b4=27,則d=a14故an=a1+(n-1)d=1+2(n-1)=2n-1.(2)由(1)知,cn=an+bn=2n-1+3n-1,所以Tn=[1+3+…+(2n-1)]+(1+3+9+…+3n-1)=12n·2n+1-3n1實(shí)力提升9.數(shù)列1,11+2,11+2+3,…,(A)2n2(C)n+2n解析:該數(shù)列的通項為an=2n(n+1),分裂為兩項差的形式為an=2(1n則Sn=2(1-12+12-13+13-14+…+所以Sn=2(1-1n+1)=10.已知函數(shù)f(n)=n2(當(dāng)n為奇數(shù)時),-n2(當(dāng)n為偶數(shù)時),且an解析:a1+a2+a3+…+a100=[f(1)+f(2)]+[f(2)+f(3)]+…+[f(100)+f(101)]=(12-22)+(-22+32)+(32-42)+…+(-1002+1012)=-3+5-7+9-…-199+201=2×50=100.答案:10011.(2024·陜西質(zhì)檢)已知正項數(shù)列{an}是首項為2的等比數(shù)列,且a2+a3=24.(1)求數(shù)列{an}的通項公式;(2)設(shè)bn=2n3an,求數(shù)列{b解:(1)設(shè)正項數(shù)列{an}的公比為q,則2q+2q2=24,所以q=3(q=-4舍去),所以an=2×3n-1.(2)因為bn=2n3an=所以Tn=13+232+333+所以13Tn=132+233+…+由①-②,得23Tn=13+132+133+所以Tn=32[13(1=3n探究創(chuàng)新12.(2024·襄陽高二檢測)已知數(shù)列{an}中,a1=1,an·an+1=(12)n,記T2n為{an}的前2n項的和,bn=a2n+a2n-1,n∈N*(1)推斷數(shù)列{bn}是否為等比數(shù)列,并求出{bn};(2)求T2n.解:(1)因為an·an+1=(12)n所以an+1·an+2=(12)n+1所以an+2a即an+2=12an因為bn=a2n+a2n-1,所以bn+1bn=a2所以{bn}是公比為12因為a1=1,a1·a2=12所以a2=12?b1=a1+a2=3所以bn=32×(12)n-1=(2)