等差数列的题已知数列bn=（a1+2a2+……+nan）/（1+2+……+n）,数列bn是等差数列,求证数列an是等差数

n=(a1+2a2+...+nan)/(1+2+...+n)
a1+2a2+...+nan=(1+2+...+n)bn=n(n+1)bn/2 (1)
a1+2a2+...(n-1)an=n(n-1)b(n-1)/2 (2)
(1)-(2)
nan=n(n+1)bn/2 -n(n-1)b(n-1)/2
an=(n+1)bn/2 -(n-1)b(n-1)/2
a(n+1)=(n+2)b(n+1)/2-nbn/2
数列{bn}是等差数列时设公差为d
an=(n+1)bn/2 -(n-1)b(n-1)/2=(n+1)bn/2 -(n-1)(bn -d)/2=bn+ (n-1)d/2
a(n+1)=(n+2)b(n+1)/2-nbn/2=(n+2)(bn +d)/2 -nbn/2=bn +(n+2)d/2
a(n+1)-an=bn+(n+2)d/2 -bn -(n-1)d/2=(3/2)d为定值
数列{an}是等差数列