Question

一个数列{an},当n为奇数时,an=5n+1,当n为偶数时,an=2^(n

Answer 1

1.奇数项：(5*1+1)+(5*3+1)+...+[5*(2m-1)+1]

=5*(1+3+...+2m-1)+m

=5m^2+m

偶数项：[2^(2/2)+2^(4/2)+...+2^(2m/2)]

=(2^1+2^2+...2^m)

=2^(m+1)-2

相加得：原式＝5m^2+2^(m+1)+2m-2

2.因为3t*Sn-(2t+3)S(n-1)=3t,3t*[S(n-1)+an]-(2t+3)S(n-1)=3t

所以(t-3)S(n-1)+3tan=3t ①

(t-3)Sn+3ta(n+1)=3t ②

②-①得

(t-3)[Sn-S(n-1)]+3t[a(n+1)-an]=0=

(t-3)(an)+3t[a(n+1)-an]=0

所以a(n+1)/an=(2t+3)/3t

即{an}是等比数列.