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。課堂10分鐘達(dá)標(biāo)1.在等比數(shù)列{an}中,若a1+a5=34,a5-a1=30,則a3=() D.±8【解析】選A.由a1+由a5=a1·q4,得q2=4,所以a3=a1·q2=8.2.在等比數(shù)列{an}中,已知其前n項(xiàng)和Sn=2n+1+a,則a的值為() 【解析】選C.當(dāng)n=1時(shí),a1=S1=22+a=4+a,當(dāng)n≥2時(shí),an=Sn-Sn-1=(2n+1+a)-(2n+a)=2n,因?yàn)閧an}為等比數(shù)列,所以a1也應(yīng)該符合an=2n,從而可得4+a=2?a=-2.3.(2023·菏澤高二檢測(cè))設(shè)Sn為等比數(shù)列an的前n項(xiàng)和,已知3S3=a4-2,3S2=a3-2,則公比q=________【解析】3S3=a4-2,3a1(1+q+q2)=a1q3-2(1),因?yàn)?S2=a3-2,所以3a1(1+q)=a1q2-2(2),由(1)(2)得:q=4,a1=2.答案:44.12+222+323【解析】設(shè)Sn=12+222+3則12Sn=122+223相減得12Sn=12+122=121-12n1-1所以Sn=2-12n-1-所以原式=-12n-1-n2答案:-n5.等比數(shù)列{an}的公比q<0,已知a2=1,an+2=an+1+2an,則S2023=________.【解析】由an+2=an+1+2an,得qn+1=qn+2qn-1,即q2-q-2=0,又q<0,解得q=-1.又a2=1,所以a1=-1,則S2023=-1×[1-(-1答案:06.數(shù)列{an}的前n項(xiàng)和為Sn,若a1=1,an+1=3Sn(n≥1),求an.【解析】當(dāng)n≥1時(shí),an+1=3Sn,則an+2=3Sn+1,所以an+2-an+1=3Sn+1-3Sn=3an+1,即an+2=4an+1,所以數(shù)列從第2項(xiàng)起是以4為公比的等比數(shù)列,又a2=3S1=3a1=3,所以an=17.【能力挑戰(zhàn)題】已知等比數(shù)列{an}中,a1=2,a3+2是a2和a4的等差中項(xiàng).(1)求數(shù)列{an}的通項(xiàng)公式.(2)記bn=anlog2an,求數(shù)列{bn}的前n項(xiàng)和Sn.【解析】(1)設(shè)數(shù)列{an}的公比為q,由題意知:2(a3+2)=a2+a4,所以q3-2q2+q-2=0,即(q-2)(q2+1)=0.所以q=2,即an=2·2n-1=2n.(2)bn=n·2n,所以Sn=1·2+2·22+3·23+…+n·2n.①2Sn=1·22+2·23+3·24+…+(n-1)·2n+n·2n+1.②①-②得-Sn=21+22+23+2