1.写出bn和bn-1的表达式,把分母乘过去,两式相减,得到
（1+2+……+n-1）(bn-bn-1)=n(an-bn)
n(n-1)(bn-bn-1)/2=n(an-bn)
即an-bn=(n-1)(bn-bn-1)/2
带入bn=b1+(n-1)d
得an=b1+(n-1)(3d/2),等差
2.代入an=a0+nd
bn=[a0+d+2a0+2*2d+……+na0+n*nd]/[1+……+n]
=a0+d*[1^2+……+n^2]/[1+……+n]
=a0+d*[n(n+1)(2n+1)/6]/[n(n+1)/2]
=a0+d(2n+1)/3
显然也是等差数列