高三数学高考复习课本回扣训练 课件六 课件VIP免费

1.已知数列{an}的前n项和Sn=n2-9n(n∈N*),第k项满足5＜ak＜8,则k等于()A.9B.8C.7D.6解析因a1=S1=-8,而当n≥2时,由an=Sn-Sn-1求得an=2n-10,此式对于n=1也成立.要满足5＜ak＜8,只须5＜2k-10＜8,从而有而k为自然数.因而只能取k=8.回扣练习六,9215kB2.如果a1,a2,…,a8为各项都大于零的等差数列,公差d≠0,则()A.a1a8＞a4a5B.a1a8＜a4a5C.a1+a8＞a4+a5D.a1a8=a4a5解析a8=a1+7d,a4=a1+3d,a5=a1+4d,∴a1·a8=+a1·7d=+7a1d,a4a5=(a1+3d)(a1+4d)=+7a1d+12d2.又∵d≠0,∴d2＞0,∴a1·a8＜a4·a5.3.已知数列{an}对于任意的p,q∈N*满足ap+q=ap+aq,且a2=-6,则a10等于()A.-165B.-33C.-30D.-21解析依题意知:a10=a2+8=a2+a8=a2+a4+a4=5a2=-30.21a21a21aBC4.在正项等比数列{an}中,若a2·a4·a6·a8·a10=32,等于()A.B.C.D.解析由a2·a4·a6·a8·a10=32,得a6=2,有等比数列的性质得:8272log21logaa81614121),(loglog21log8728272aaaa.21log)(loglog21log,628728272867aaaaaaaaD5.已知等比数列a1,a2,a3的和为定值m(m＞0),且其公比q＜0,令t=a1a2a3,则t的取值范围是()A.(0,m3]B.[-m3,0)C.[-m3,m3]D.[-m3,0)∪(0,m3]解析因为m=a1+a2+a3=所以因为q＜0,所以(当且仅当q=-1时取等号).又m＞0,所以故t=a1a2a3=),11(2qqa,112qqma21qq,0,0112amqqmm即).0,[332maB6.已知等差数列{an}中,an≠0,若m＞1,且am-1-+am+1=0,S2m-1=38,则m的值为()A.38B.20C.19D.10解析由{an}为等差数列得am-1+am+1=2am(m＞1,m∈N*),又这里am-1-+am+1=0,故得则这里am≠0,∴am=2，再由S2m-1=38,得得2m-1=19,解得m=10.2ma2ma,22mmaa,382)12)((121maamD7.已知等比数列的公比为2,且前4项之和等于1,那么它的前8项之和等于_____.解析由已知得S4=1,q=2,8.设{an}是公比为q的等比数列,Sn是它的前n项和,若{Sn}是等差数列,则q=___.解析注意到又{Sn}为等差数列,∴当n≥2时,an=Sn-Sn-1=…=S2-S1=a2,∴a1qn-1=a1q,而a1≠0,∴qn-2=1,即q=1.,121)21(41a.1721)21(151,151881Sa即,)2()1(11nSSnSannn1719.各项都是正数的等比数列{an}的公比q≠1,且a2,a1成等差数列,则=_________.解析注意到只要求出q;由已知条件得∴a1q2=a1(1+q)q2-q-1=0,由此解得q=∵an＞0,∴q＞0,∴q=于是得,213a5443aaaa,1)(43435443qaaqaaaaaa,)21(2213aaa,251,251.2155125443aaaa21510.设数列{an}的前n项和为Sn,(n≥1),且a4=54,则a1=____.解析由已知得a4=S4-S3=∴a1(81-1)-a1(27-1)=108,∴54a1=108,∴a1=2.11.证明:若f(x)=ax+b,且{xn}是等差数列,则{f(xn)}也是等差数列.证明已知{xn}是等差数列,设任意n∈N，有xn+1-xn=r(r是公差).于是,任意n∈N,有f(xn+1)-f(xn)=(axn+1+b)-(axn+b)=a(xn+1-xn)=ar,其中ar是常数,即{f(xn)}也是等差数列.2)13(1nnaS,542)13(2)13(3141aa212.已知数列{an}的前n项和Sn=n2-48n.(1)求数列的通项公式；(2)求Sn的最大或最小值.解(1)a1=S1=12-48×1=-47,当n≥2时,an=Sn-Sn-1=n2-48n-[(n-1)2-48(n-1)]=2n-49,a1也适合上式，∴an=2n-49(n∈N+).(2)a1=-49,d=2,所以Sn有最小值,∴n=24,即Sn最小,或:由Sn=n2-48n=(n-24)2-576,∴当n=24时,Sn取得最小值-576.,,21242123,049)1(204921Nnnnanann又得由,576222324)47(2424S返回