Question

数列{an}的前n项和为SN,且a1=1.an+1=1/3sn,n=1,2,3...

求(1)数列{an}通项公式 （2）求a1+a2+a3+..+a2n-1的值

Answer 1

a(n+1)
=
sn/3,
an
=
s(n-1)/3,
两式相减得
a(n+1)
-
an
=
[sn
-
s(n-1)]/3
=
an/3
a(n+1)
=
4an/3
a(n+1)
/an=4/3
{an}
是首项为1，公比为4/3的等比数列，
an=a1q^(n-1)
an
=
(4/3)^(n-1),
n
=
1,2,...
a1+a2+a3+..+a2n-1
=[1-(4/3)^(2n-1)]/(1-4/3)
=3[(4/3)^(2n-1)-1]
=3*(4/3)^(2n-1)-3

Answer 2

a(n)=s(n)-s(n-1)=3a(n+1)-3a(n)
(n>=2)
所以4a(n)=3a(n+1)
所以a(n+1)=4/3*a(n)
所以a(n)为公比为4/3的等比数列(n>=2)
所以a(n)=1
(n=1)
a(n)=(4/3)^(n-2)*1/3
(n>=2)
a2+a4+...+a2n
=1/3+1/3*(4/3)^2+1/3*(4/3)^4+
...+1/3*(4/3)^(2n-2)
=1/3*((16/9)^0+(16/9)^1+(16/9)^2+...+(16/9)^(n-1))
=1/3*(1-(16/9)^n)/(1-16/9)=3/7*((16/9)^n-1)