设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).(1)令bn=an+1-an(n=1,2''''''),求数列(bn)的通项公式.(2)求数列(an)的前n项和Sn (备注题中的列如:an+2中的n+2为a的底,其它

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设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).(1)令bn=an+1-an(n=1,2''''''),求数列(bn)的通项公式.(2)求数列(an)的前n项和Sn (备注题中的列如:an+2中的n+2为a的底,其它

设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).(1)令bn=an+1-an(n=1,2''''''),求数列(bn)的通项公式.(2)求数列(an)的前n项和Sn (备注题中的列如:an+2中的n+2为a的底,其它
设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).
(1)令bn=an+1-an(n=1,2''''''),求数列(bn)的通项公式.(2)求数列(an)的前n项和Sn
(备注题中的列如:an+2中的n+2为a的底,其它的an+1也如此)

设数列〔an〕满足a1=1,a2=5/3(3分之5),an+2=5/3an+1-2/3an,(n属于N※).(1)令bn=an+1-an(n=1,2''''''),求数列(bn)的通项公式.(2)求数列(an)的前n项和Sn (备注题中的列如:an+2中的n+2为a的底,其它
(1)a(n+2)=(5/3)*a(n+1)-(2/3)*a(n)
3a(n+2)=5a(n+1)-2*a(n)
3a(n+2)-3a(n+1)=2a(n+1)-2a(n)
3[a(n+2)-a(n+1)]=2[a(n+1)-a(n)]
[a(n+2)-a(n+1)]/[a(n+1)-a(n)]=2/3
即b(n+1)/b(n) =2/3
b(n)是等比数列
b1=a2-a1=5/3-1=2/3
b(n)是首项为2/3,公比为2/3的等比数列
b(n)=(2/3)^n
(2)an-a(n-1)=b(n-1)=(2/3)^(n-1)
a(n-1)-a(n-2)=b(n-2)=(2/3)^(n-2)
.
a3-a2=b2=(2/3)^2
a2-a1=b1=(2/3)^1
各等式相加得
an-a1=(2/3)^1+(2/3)^2+…+(2/3)^(n-1)
an=a1+(2/3)*(1-(2/3)^(n-1))/(1-2/3)
=1+2(1-(2/3)^(n-1))
=3-2(2/3)^(n-1)
Sn=3n-2[(1-(2/3)^n)/(1-2/3)]=3n+6(2/3)^n-6

拜托,别拿这些问题放到这上面!看到了很烦的啦……

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