设两个数列{a},{b}满足dn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n),若{bn}为等差数列,求证:{an}也为等差数列

Question

设两个数列{a},{b}满足dn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n),若{bn}为等差数列,求证:{an}也为等差数列

Answer 1

[n(n+1)/2]bn=a1+2a2+3a3+…+nan,所以[(n-1)n/2]b(n-1)=a1+2a2+3a3+…+(n-1)a(n-1),上述两式相减得,[n(n+1)/2]bn-[(n-1)n/2]b(n-1)=nan,两边同除以n得,[（n+1)/2]bn-[(n-1/2)]b(n-1)=an,所以（1/2）nbn-1/2nb(n-1)+(1/2)bn+(1/2)b(n-1)=an,{bn}为等差数列,因此设bn=b1+(n-1)d,代入（1/2）nbn-1/2nb(n-1)+(1/2)bn+(1/2)b(n-1)=an,得an=(1/2)nd+(1/2)(2b1+2nd-3d)=b1+(n-1)(3/2)d,
因此若bn是公差为d 的等差数列,则an为是公差为（3/2)d的等差数列.