数列{ an}{ bn}满足关系式bn=1*a1+2*a2+3*a3…+nan/1+2+3+…+n,若{bn}为等差数列,求证数列{an}也是等差数
问题描述：
数列{ an}{ bn}满足关系式bn=1*a1+2*a2+3*a3…+nan/1+2+3+…+n,若{bn}为等差数列,求证数列{an}也是等差数
答
设bn = xn + y
则
an = { n(n-1)/2*(xn+y) - n(n+1)/2*[x(n-1)+y] }/n
= (n-1)/2*(xn+y) - (n+1)/2*[x(n-1)+y]
= 1/2 * [ -nx + (x-2y) ]
因此，an也为等差数列
答
证：设{bn}公差为d (d为常数).a1/1=b1 a1=b1bn=(a1+2a2+3a3+...+nan)/(1+2+3+...+n)a1+2a2+3a3+...+nan=[n(n+1)/2]bn (1)a1+2a2+3a3+...+(n+1)a(n+1)=[(n+1)(n+2)/2]b(n+1) (2)(2)-(1)(n+1)a(n+1)=[(n+1)(n+2)/2]b(n...