已知:bn=(a1+2a2+...+nan)/(1+2+...+n),数列an成等差数列的充要条件是bn也是等差数列.

问题描述:

已知:bn=(a1+2a2+...+nan)/(1+2+...+n),数列an成等差数列的充要条件是bn也是等差数列.

bn=(a1+2a2+...+nan)/(1+2+...+n)
a1+2a2+...+nan=(1+2+...+n)bn=n(n+1)bn/2 (1)
a1+2a2+...(n-1)an=n(n-1)b(n-1)/2 (2)
(1)-(2)
nan=n(n+1)bn/2 -n(n-1)b(n-1)/2
an=(n+1)bn/2 -(n-1)b(n-1)/2
a(n+1)=(n+2)b(n+1)/2-nbn/2
1.
数列{an}是等差数列时,a(n+1)-an为定值.
(n+2)b(n+1)/2 -nbn/2 -(n+1)bn/2 +(n-1)b(n-1)/2
=[[b(n+1)-2bn+b(n-1)/2]n +[2b(n+1)-bn-b(n-1)]/2
要对任意正整数n,[[b(n+1)-2bn+b(n-1)/2]n +[2b(n+1)-bn-b(n-1)]/2恒为定值,则只有n项系数=0
[b(n+1)-2bn+b(n-1)]/2=0
b(n+1)-2bn+b(n-1)=0
b(n+1)-bn=bn-b(n-1)
数列{bn}是等差数列.
2.
数列{bn}是等差数列时,设公差为d
an=(n+1)bn/2 -(n-1)b(n-1)/2=(n+1)bn/2 -(n-1)(bn -d)/2=bn+ (n-1)d/2
a(n+1)=(n+2)b(n+1)/2-nbn/2=(n+2)(bn +d)/2 -nbn/2=bn +(n+2)d/2
a(n+1)-an=bn+(n+2)d/2 -bn -(n-1)d/2=(3/2)d,为定值.
数列{an}是等差数列.
综上,得数列an成等差数列的充要条件是bn也是等差数列.