Question

已知数列an是首项与公比均为1/3的等比数列,数列bn的前n项和Bn=1/2（n²+n）

Answer 1

an=(1/3)^n, Bn=n(n+1)/2, bn=Bn-Bn-1=n(n+1)/2-(n-1)n/2=n

anbn=(1/3)^n*n

Sn=1/3*1+(1/3)^2*2+(1/3)^3*3+.+(1/3)^(n-1)*(n-1)+(1/3)^n*n (1)

(1)式两边乘以1/3

1/3Sn= (1/3)^2*1+(1/3)^3*2+(1/3)^4*3+ .+(1/3)^n*(n-1)+(1/3)^(n+1)*n (2)

(1)式-(2)式,得

2/3Sn=[1/3+(1/3)^2+(1/3)^3+.+(1/3)^n]-(1/3)^(n+1)*n

=1/3[1-(1/3)^n]/(1-1/3)-(1/3)^(n+1)*n

=1/2-(1/2+n/3)(1/3)^n

Sn=3/4-(3/4+n/2)(1/3)^n=1/3

所以1/3