新版高中數(shù)學(xué)北師大版必修5習(xí)題第一章數(shù)列1.2.2.1

2.2等差數(shù)列的前n項(xiàng)和第1課時(shí)等差數(shù)列的前n項(xiàng)和課時(shí)過關(guān)·能力提升1.等差數(shù)列{an}的各項(xiàng)都是負(fù)數(shù),且a32+a82+2a3a8=9,則它的前10項(xiàng)和A.11 B.9 C.15 D.13解析:∵a32+a82+2∴(a3+a8)2=(a1+a10)2=9.∵an<0,∴a1+a10<0.∴a1+a10=3,∴S10=10(a1答案:C2.設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,若滿足an=an1+2(n≥2),且S3=9,則a1等于()A.5 B.3 C.1 D.1解析:∵an=an1+2(n≥2),∴anan1=2(n≥2),∴{an}是公差為2的等差數(shù)列.∵S3=a1+a2+a3=3a2=9,∴a2=3.∴a1=a2d=32=1.答案:C3.若等差數(shù)列{an}滿足a5=11,a12=3,{an}的前n項(xiàng)和Sn的最大值為M,則lgM等于()A.1 B.2 C.10 D.100答案:B4.在等差數(shù)列{an}中,a1+a4+a7=39,a3+a6+a9=27,則數(shù)列{an}的前9項(xiàng)和S9等于()A.66 B.99 C.144 D.297解析:由a1+a4+a7=39,a3+a6+a9=27,得3a4=39,3a6=27,解得a4=13,a6=9,所以S9=9(a1答案:B5.在等差數(shù)列{an}中,a9=12a12+6,則數(shù)列{an}的前11項(xiàng)和S11等于(A.24 B.48 C.66 D.132解析:設(shè)等差數(shù)列{an}的公差為d,則由a9=12a12+6,得a1+8d=12(a1+11d)+6,整理得a1+5d=12,即a6=12,所以S11=11(a1+a11)2=答案:D6.{an}為等差數(shù)列,a10=33,a2=1,Sn為數(shù)列{an}的前n項(xiàng)和,則S202S10等于()A.40 B.200 C.400 D.20解析:S202S10=20(a1+a20)22×10(a1∵a10=a2+8d,∴33=1+8d,∴d=4.∴S202S10=400.答案:C★7.設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,且滿足S15>0,S16<0,則S1a1,S2a2A.S1a1 BC.S8a8解析:∵S15>0,∴a1+a15=2a8>0,∴a8>0.∵S16<0,∴a1+a16<0,∴a8+a9<0,∴a9<0,∴S8最大.又a1>a2>a3>…>a8>0>a9>…,∴S1a1,S2a答案:C8.已知數(shù)列{an}的前n項(xiàng)和Sn=n2+n,則它的通項(xiàng)公式為an=.

解析:當(dāng)n=1時(shí),a1=S1=2;當(dāng)n≥2時(shí),an=SnSn1=(n2+n)[(n1)2+(n1)]=2n.∵a1=2也符合上式,∴an=2n.答案:2n9.設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若a1=11,a4+a6=6,則當(dāng)Sn取最小值時(shí),n等于.

答案:610.設(shè)Sn為等差數(shù)列{an}的前n項(xiàng)和,若S3=3,S6=24,則a9=.

解析:設(shè)等差數(shù)列{an}的公差為d,則S3=3a1+3×22d=3a1+即a1+d=1.①S6=6a1+6×52d=6a1+即2a1+5d=8.②聯(lián)立①②兩式,解得a1=1,d=2,故a9=a1+8d=1+8×2=15.答案:1511.已知數(shù)列{an}的前n項(xiàng)和為Sn,且滿足log2(Sn+1)=n+1,求數(shù)列{an}的通項(xiàng)公式.解:由已知條件,得Sn+1=2n+1,則Sn=2n+11.所以當(dāng)n=1時(shí),a1=S1=3;當(dāng)n≥2時(shí),an=SnSn1=(2n+11)(2n1)=2n.又當(dāng)n=1時(shí),3≠21,故an=3★12.甲、乙兩物體分別從相距70m的兩處同時(shí)相向運(yùn)動(dòng),甲第1分鐘走2m,以后每分鐘比前1分鐘多走1m,乙每分鐘走5m.(1)甲、乙開始運(yùn)動(dòng)幾分鐘后第一次相遇?(2)如果甲、乙到達(dá)對(duì)方起點(diǎn)后立即折返,甲繼續(xù)每分鐘比前1分鐘多走1m,乙繼續(xù)每分鐘走5m,那么開始運(yùn)動(dòng)幾分鐘后第二次相遇?解:(1)設(shè)甲、乙開始運(yùn)動(dòng)nmin后第一次相遇,依題意,有2n+n(n-1)整理得n2+13n140=0,解得n=7或n=20(舍去).故甲、乙開始運(yùn)動(dòng)7