{an}和{bn}满足bn=(a1+2a2+.+nan)/(1+2+.+n) 求证:(1)若{bn}为等差数列,{an}也是等差数列

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{an}和{bn}满足bn=(a1+2a2+.+nan)/(1+2+.+n)求证:(1)若{bn}为等差数列,{an}也是等差数列{an}和{bn}满足bn=(a1+2a2+.+nan)/(1+2+.

{an}和{bn}满足bn=(a1+2a2+.+nan)/(1+2+.+n) 求证:(1)若{bn}为等差数列,{an}也是等差数列
{an}和{bn}满足bn=(a1+2a2+.+nan)/(1+2+.+n) 求证:(1)若{bn}为等差数列,{an}也是等差数列

{an}和{bn}满足bn=(a1+2a2+.+nan)/(1+2+.+n) 求证:(1)若{bn}为等差数列,{an}也是等差数列
设{bn}公差为d
bn=(a1+2a2+.+nan)/(1+2+.+n)=(a1+2a2+.+nan)/[n(n+1)/2]
[n(n+1)/2]bn=a1+2a2+.+(n-1)a(n-1)+nan-------------(1)
[n(n-1)/2]b(n-1)=a1+2a2+.+nan-----------------------(2)
其中a(n-1)和b(n-1)分别表示{an},{bn}的第n-1项.
(1)-(2)得
[n(n+1)/2]bn-[n(n-1)/2]b(n-1)=nan
即:(n^2/2)[bn-b(n-1)]+(n/2)[bn+b(n-1)]=nan
所以an=(n/2)d+(1/2)[bn+b(n-1)]
a(n+1)-an=[(n+1)/2]d+(1/2)[b(n+1)+bn]-(n/2)d-(1/2)[bn+b(n-1)]
=(1/2)d+(1/2)[b(n+1)-b(n-1)]
=(1/2)d+(1/2)*2d
=(3/2)d
所以a(n+1)-an为常数
所以{an}是等差数列

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