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《高考数学第一轮复习15数列的通项》由会员上传分享,免费在线阅读,更多相关内容在学术论文-天天文库。
1、15.数列的通项班级姓名一.选择题:1.已知数列{an}的前四项依次是:20,11,2,7,那么它的一个通项公式是()(A)an=9n+11(B)an=-9n+20(C)(D)2.在数列{an}中,an+1=an+n,且a1=1,则a100=()(A)4989(B)4949(C)4950(D)49513.如果数列{an}的前n项和Sn=an-3,那么这个数列的通项公式是( )(A)an=2(n2+n+1)(B)an=3×2n(C)an=3n+1(D)an=2×3n4.数列{an}的首项a1=-1,anan+1=3,则其通项公式是( )(A)an=2+
2、(-1)n(B)an=2+(-1)n+1(C)an=-2+(-1)n(D)an=-2+(-1)n+15.若Sn是数列{an}的前n项和,且Sn=n2,则{an}是( )(A)等比数列,但不是等差数列(B)等差数列,但不是等比数列(C)等差数列,而且也是等比数列(D)既非等差数列又非等比数列二.填空题:6.数列,的一个通项公式是.7.已知数列{an}中,a1=,当n2时,有,则{an}的通项公式为.8.在数列{an}中,a1=6,an=2an-1-4,(n≥2),则数列的通项公式an=.9.设{an}是首项为1的正项数列,且,(n=1,2,3,…),则它
3、的通项公式an=.三.解答题:10.设{an}是一个公差为d(d0)的等差数列,它的前10项和S10=110,且a1,a2,a4成等比数列.(1)证明:a1=d(2)求公差d的值和数列{an}的通项公式.11.设二次函数f(x)=x2+x,当x∈[n,n+1](nN﹡)时,f(x)的所有整数值的个数为g(n)(1)求g(n)的表达式;(2)设(nN﹡),Sn=a1-a2+a3-a4+…+(-1)n-1an,求Sn;(3)设,Tn=b1+b2+…+bn,若Tn4、k+1=a2k+3k,其中k=1,2,3,…(1)求a3,a5(2)求{an}的通项公式.内部资料仅供参考9JWKffwvG#tYM*Jg&6a*CZ7H$dq8KqqfHVZFedswSyXTy#&QA9wkxFyeQ^!djs#XuyUP2kNXpRWXmA&UE9aQ@Gn8xp$R#͑Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849G5、x^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5ux^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vST6、T#&ksv*3tnGK8!z89AmUE9aQ@Gn8xp$R#͑Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5ux^Gjqv^$UE9wEwZ#Q7、c@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z8vG#tYM*Jg&6a*CZ7H$dq8KqqfHVZFedswSyXTy#&QA9wkxFyeQ^!djs#XuyUP2kNXpRWXmA&UE9aQ@Gn8xp$R#͑Gx^Gjq8、v^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm
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