设{Bn}公差为d
An=1*B1+2*B2+3*B3+...n*Bn
=B1+2(B1+d)+3(B1+2d)+4(B1+3d)+...+n[B1+(n-1)]d
=B1+2B1+3B1+...+nB1 + 2d+6d+12d+...+(n-1)nd
=B1*(1+2+3+...+n) + d*[1*2+2*3+...+(n-1)n]
=B1*n(n+1)/2 + d*(n-1)n(n+1)/3
因为 An=n(n+1)^2
所以 B1*n(n+1)/2 + d*(n-1)n(n+1)/3＝n(n+1)^2
两边同时除以n(n+1),得：B1/2 + (n-1)d/3 = n+1
3B1 + 2(n-1)d = 6n+6
B1 + 2B1 + 2(n-1)d = 6n+6
B1 + 2Bn = 6n+6
所以 Bn=3n+1
即数列{Bn}是以4为首项,3为公差的等差数列