已知数列{an}和{bn}满足:bn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n)求当{an}是等差数列的时候证明{bn}是等差数

设an公差为d,则bn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n)=2(a1+2a2+3a3+…+nan)/n(n+1)=2(a1+2(a1+d)+3(a1+2d)+…+n(a1+(n-1)d)/n(n+1)=2{(a1+2a1+3a1+…+na1)+[1*2+2*3+3*4+…(n-1)n]d}/n(n+1)=2{(n(n+1)a1/2)+[1*2+2*3...