1.設< an >為等比級數,已知S3+S6 = 2S9 則公比 = ? 2.設< an >之Sn = 2an - 1 , n = 1,2..... < bn >滿足 b1 = 3, bk+1 = ak + bk k = 1,2..... 求< bn >的前n項和高中 回應 bubupin Re: 數列級數 2007-11-02 21:53 1.設< an >為等比級數,已知S3+S6 = 2S9 則公比 = ? 設公比為r 2(a+ar+ar^2)+ar^3+ar^4+ar^5=2(a+ar+ar^2+........ar^8) ar^3+ar^4+ar^5=-2(ar^6+ar^7+ar^8) ar^3(1+r+r^2)=-2ar^6(1+r+r^2) ar^3不等於0,r^2+r+1 > 0,故r^3=-1/2,r=(-1/2)^(1/3) 2.設< an >之Sn = 2an - 1 , n = 1,2..... < bn >滿足b1 = 3,bk+1 = ak + bk,k = 1,2..... 求< bn >的前n項和 Sn-Sn-1=an=2(an-an-1) an=2an-1 故< an > 為首項1,公比2之等比數列 b2-b1=a1 b3-b2=a2 .............. bn-bn-1=an-1 bn-b1=2^(n-1) - 1,bn=2^(n-1)+2 b1+.................bn=2^n+2n-1 路過 Re: 數列級數 2007-11-02 22:05 大哥= =你算錯了喔 應該是bn-b1=2^(n-1) - 1,bn=2^(n-1)+2 bubupin Re: 數列級數 2007-11-02 22:12 謝謝
bubupin Re: 數列級數 2007-11-02 21:53 1.設< an >為等比級數,已知S3+S6 = 2S9 則公比 = ? 設公比為r 2(a+ar+ar^2)+ar^3+ar^4+ar^5=2(a+ar+ar^2+........ar^8) ar^3+ar^4+ar^5=-2(ar^6+ar^7+ar^8) ar^3(1+r+r^2)=-2ar^6(1+r+r^2) ar^3不等於0,r^2+r+1 > 0,故r^3=-1/2,r=(-1/2)^(1/3) 2.設< an >之Sn = 2an - 1 , n = 1,2..... < bn >滿足b1 = 3,bk+1 = ak + bk,k = 1,2..... 求< bn >的前n項和 Sn-Sn-1=an=2(an-an-1) an=2an-1 故< an > 為首項1,公比2之等比數列 b2-b1=a1 b3-b2=a2 .............. bn-bn-1=an-1 bn-b1=2^(n-1) - 1,bn=2^(n-1)+2 b1+.................bn=2^n+2n-1
回應
bubupin
Re: 數列級數
2007-11-02 21:53則公比 = ?
設公比為r
2(a+ar+ar^2)+ar^3+ar^4+ar^5=2(a+ar+ar^2+........ar^8)
ar^3+ar^4+ar^5=-2(ar^6+ar^7+ar^8)
ar^3(1+r+r^2)=-2ar^6(1+r+r^2)
ar^3不等於0,r^2+r+1 > 0,故r^3=-1/2,r=(-1/2)^(1/3)
2.設< an >之Sn = 2an - 1 , n = 1,2.....
< bn >滿足b1 = 3,bk+1 = ak + bk,k = 1,2.....
求< bn >的前n項和
Sn-Sn-1=an=2(an-an-1)
an=2an-1
故< an > 為首項1,公比2之等比數列
b2-b1=a1
b3-b2=a2
..............
bn-bn-1=an-1
bn-b1=2^(n-1) - 1,bn=2^(n-1)+2
b1+.................bn=2^n+2n-1
路過
Re: 數列級數
2007-11-02 22:05應該是bn-b1=2^(n-1) - 1,bn=2^(n-1)+2
bubupin
Re: 數列級數
2007-11-02 22:12