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《高考数学第一轮复习17数学归纳法及其应用》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、17.数学归纳法及其应用班级姓名一.选择题:1.利用数学归纳法证明:“(a1,nN﹡)”时,在验证n=1成立时,左边应该是 ()(A)1 (B)1+a (C)1+a+a2 (D)1+a+a2+a32.用数学归纳法证明:“,(nN﹡,n2)”的过程中,由“n=k”变到“n=k+1”时,不等式左边的变化是()(A)增加(B)增加和(C)增加,并减少(D)增加和,并减少3.若k棱柱有f(k)个对角面,则k+1棱柱有对角面的个数为()(A)2f(k)(B)k-1+f(k)(C)f(k)+k(D)f(k)+24.在证明
2、“已知,求证f(2n)3、k+1)=f(k)+.8.已知数列{an}满足a1+2a2+3a3+…+nan=4-(n+2)an,则a1,a2,a3,a4分别为,由此可猜想其通项公式an=.9.由下列各式:,可归纳出:.三.解答题10.用数学归纳法证明:11.已知数列{an}的各项均为正数,它的前n项和Sn与通项an有关系,(1)写出数列{an}的前四项;(2)猜想{an}的通项公式,并给出证明.12.设,问是否存在n的整式g(n),使得等式a1+a2+a3+…+an-1=g(n)(an-1)对于大于1的一切自然数n都成立?证明你的结论.内部资料仅供参考9JWKffwvG#tYM*Jg&6a*CZ7H$dq8K4、qqfHVZFedswSyXTy#&QA9wkxFyeQ^!djs#XuyUP2kNXpRWXmA&UE9aQ@Gn8xp$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&MuWFA5u5、x^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!z89AmUE9aQ@Gn8xp$R#͑Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z6、89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5ux^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wE7、wZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z8vG#tYM*Jg&6a*CZ7H$dq8KqqfHVZFedswSyXTy#&QA9wkxFyeQ^!djs#XuyUP2kNXpRWXmA&UE9aQ@Gn8xp$R#͑Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu#
3、k+1)=f(k)+.8.已知数列{an}满足a1+2a2+3a3+…+nan=4-(n+2)an,则a1,a2,a3,a4分别为,由此可猜想其通项公式an=.9.由下列各式:,可归纳出:.三.解答题10.用数学归纳法证明:11.已知数列{an}的各项均为正数,它的前n项和Sn与通项an有关系,(1)写出数列{an}的前四项;(2)猜想{an}的通项公式,并给出证明.12.设,问是否存在n的整式g(n),使得等式a1+a2+a3+…+an-1=g(n)(an-1)对于大于1的一切自然数n都成立?证明你的结论.内部资料仅供参考9JWKffwvG#tYM*Jg&6a*CZ7H$dq8K
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