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_数学解答

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

人气：213 ℃　时间：2019-08-18 21:01:02

解答

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...