Question

已知数列｛an｝的前n项和Sn=a[2-(1/2)^n-1]-b[2-(n+1)(1/2)^n-1]其中a,b是非零常数，则数列{xn}{yn}使得

A、an=xn+yn,其中{xn}为等差数列，{yn}为等比数列
B、an=xn+yn,其中{xn}和{yn}都为等差数列
C、an=xn*yn,其中{xn}为等差数列，{yn}为等比数列
D、an=xn*yn,其中{xn}和{yn}为等比数列

Answer 1

C
Sn=a[2-(1/2)^n-1]-b[2-(n+1)(1/2)^n-1]
Sn-1=a[2-(1/2)^n-2]-b[2-n(1/2)^n-2]
an=Sn-Sn-1=a{[2-(1/2)^n-1]-[2-(1/2)^n-2]}-b{[2-(n+1)(1/厅桥2)^n-1]-[2-n(1/2)^n-2]}
=a*(1/2)^(n-1)-b(n-1)/[2^(n-1)]
=(1/2)^(n-1)[a-b(n-1)]
设yn=(1/源租2)^(n-1)
xn=a-b(n-1)
则yn为扮裂猛等比数列
xn为等差数列