bn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n),
n(n+1)bn=2(a1+2a2+3a3+…+nan)
(n+1)(n+2)b=2[a1+2a2+3a3+…+nan+(n+1)a]
=n(n+1)bn+2(n+1)a
(n+2)b=nbn+2a
设{bn}公差d
n(b-bn)+2b=2a
nd+2b=2a (1)
(n-1)d+2bn=2an (2)
(1)-(2)
d+2d=2(a-an)
a-an=(3/2)d为常数
检验,b1=a1,b2=(a1+2a2)/3
b2-b1=(2/3)(a2-a1)
a2-a1=(3/2)(b2-b1)=(3/2)d成立
所以数列{an}成等差数列的充要条件是数列{bn}也是等差数列
答：设An=a+(n-1)d B1=2a+d B2=2a+(1+2)d=2a+3d B3=2a+(2+3)d=2a+5d .................. Bn...详情>>
答：详情>>
