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(留存版)
【正文】 016.答案2 016{an}滿足a1=1,an+1解析當(dāng)n≥2時(shí),an=SnSn1=(b1)(2)設(shè)數(shù)列{an+bn}是首項(xiàng)為1,公比為|a2|的等比數(shù)列,求{bn}的前n項(xiàng)和Sn.解(1)設(shè)等差數(shù)列{an}的公差為d,依題意得a3+a8(a2+a7)=2d=6,從而d=3.所以a2+a7=2a1+7d=23,解得a1=1.所以數(shù)列{an}的通項(xiàng)公式為an=3n+2.(2)由(1)得a2=4,所以|a2|=4.而數(shù)列{an+bn}是首項(xiàng)為1,公比為4的等比數(shù)列.所以an+bn=4n1,即3n+2+bn=4n1,所以bn=3n2+4n1,于是Sn=[1+4+7+…+(3n2)]+(1+4+42+…+4n1)=n(3n1)2+14n14=n(3n1)2+4n13.{an}的前n項(xiàng)和為Sn,且a1=1,an+1=Sn,n∈N*,求:(1)a2,a3,a4的值及數(shù)列{an}的通項(xiàng)公式。21 0093.答案3(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,
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