數(shù)列252等比數(shù)列前n項(xiàng)和的性質(zhì)及應(yīng)用練習(xí)新人教a版必修5-資料下載頁(yè)
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【正文】 {an}的公差不為0,a2,a5,a14恰好是等比數(shù)列{bn}的前三項(xiàng),a2=3.(1)求數(shù)列{an},{bn}的通項(xiàng)公式。(2)記數(shù)列{bn}的前n項(xiàng)和為Tn,若對(duì)任意的n∈N*,kTn+32≥3n6恒成立,求實(shí)數(shù)k的取值范圍.解(1)設(shè)公差為d,根據(jù)題意知d≠0,a2=a1+d,a5=a1+4d,a14=a1+13d.∵(a1+4d)2=(a1+d)(a1+13d),a1+d=3,∴3d26d=0,∴d=2(d=0舍去).又a2=3,d=2,∴a1=1,an=2n1.∵b1=a2=3,b2=a5=9,b3=a14=27,∴bn=3n.(2)由(1)知b1=3,q=3.∵Tn=b1(1qn)1q=3(13n)13=3n+132,∴3n+132+32k≥3n6對(duì)n∈N*恒成立.∴Tn0,∴k≥2n43n對(duì)n∈N*恒成立.令=2n43n,1=2n43n2n63n1=2(2n7)3n,當(dāng)n≤3時(shí),1,當(dāng)n≥4時(shí),1,∴()max=c3=227,故k≥227.{an}的前n項(xiàng)和為Sn,且a2=8,S4={bn}的前n項(xiàng)和為Tn,且Tn2bn+3=0,n∈N*.(1)求數(shù)列{an},{bn}的通項(xiàng)公式。(2)設(shè)=an,n為奇數(shù),bn,n為偶數(shù),求數(shù)列{}的前2n+1項(xiàng)和P2n+1.解(1)由題意知,a1+d=8,4a1+6d=40,解得a1=4,d=4,∴an=4n.∵Tn2bn+3=0,∴當(dāng)n=1時(shí),b1=3,當(dāng)n≥2時(shí),Tn12bn1+3=0,兩式相減,得bn=2bn1(n≥2),故數(shù)列{bn}為等比數(shù)列,且bn=32n1.(2)由(1)知=4n,n為奇數(shù),32n1,n為偶數(shù).∴P2n+1=(a1+a3+…+a2n+1)+(b2+b4+…+b2n)=(n+1)[4+4(2n+1)]2+6(14n)14=22n+1+4n2+8n+2.
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