设数列〔an〕满足a1=1,a2=5/3(5分之3),an+2=5/3an+1-2/3an,(n属于N※).
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设数列〔an〕满足a1=1,a2=5/3(5分之3),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)
注意有更正哈:那个a2是等于3分之5的哦~
(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)
注意有更正哈:那个a2是等于3分之5的哦~
(1)
a(n+2)=(5/3)*a(n+1)-(2/3)*a(n)
a(n+2)=(2/3)*a(n+1)+a(n+1)-(5/3)*a(n)
a(n+2)-a(n+1)=(2/3)[a(n+1)-a(n)]
即b(n+1)=(2/3)b(n)
b(n)是等比数列
b1=a2-a1=5/3-1=2/3
首项为2/3 公比为2/3
b(n)=(2/3)^n
n>=1 n是整数
(2)
a(n)-a(n-1)=b(n-1)=(2/3)^(n-1)
a(n-1)-a(n-2)=b(n-2)=(2/3)^(n-2)
.
a(3)-a(2)=b(2)=(2/3)^2
a(2)-a(1)=b(1)=(2/3)^1
各等式相加得
a(n)-a(1)=(2/3)^1+(2/3)^2+.+(2/3)^(n-2)+(2/3)^(n-1)
a(n)=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=a1+a2+...+a(n)
=3-2*1+3-2*(2/3)+...+3-2(2/3)^(n-1)
=3n-2(1+2/3+.+(2/3^(n-1)))
=3n-2(1-(2/3)^n)/(1-2/3)
=3n-6(1-(2/3)^n)
=3n+6(2/3)^n-6
a(n+2)=(5/3)*a(n+1)-(2/3)*a(n)
a(n+2)=(2/3)*a(n+1)+a(n+1)-(5/3)*a(n)
a(n+2)-a(n+1)=(2/3)[a(n+1)-a(n)]
即b(n+1)=(2/3)b(n)
b(n)是等比数列
b1=a2-a1=5/3-1=2/3
首项为2/3 公比为2/3
b(n)=(2/3)^n
n>=1 n是整数
(2)
a(n)-a(n-1)=b(n-1)=(2/3)^(n-1)
a(n-1)-a(n-2)=b(n-2)=(2/3)^(n-2)
.
a(3)-a(2)=b(2)=(2/3)^2
a(2)-a(1)=b(1)=(2/3)^1
各等式相加得
a(n)-a(1)=(2/3)^1+(2/3)^2+.+(2/3)^(n-2)+(2/3)^(n-1)
a(n)=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=a1+a2+...+a(n)
=3-2*1+3-2*(2/3)+...+3-2(2/3)^(n-1)
=3n-2(1+2/3+.+(2/3^(n-1)))
=3n-2(1-(2/3)^n)/(1-2/3)
=3n-6(1-(2/3)^n)
=3n+6(2/3)^n-6
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