7 设两个数列{an},{bn}满足bn=a1+a2+3a3+…nan/1+2+3…+n,若{bn}为等差数列,求证:{an}也为等差数列.
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![7 设两个数列{an},{bn}满足bn=a1+a2+3a3+…nan/1+2+3…+n,若{bn}为等差数列,求证:{an}也为等差数列.](/uploads/image/z/5081775-15-5.jpg?t=7+%E8%AE%BE%E4%B8%A4%E4%B8%AA%E6%95%B0%E5%88%97%7Ban%7D%2C%7Bbn%7D%E6%BB%A1%E8%B6%B3bn%EF%BC%9Da1%2Ba2%2B3a3%2B%E2%80%A6nan%2F1%2B2%2B3%E2%80%A6%2Bn%2C%E8%8B%A5%7Bbn%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%B1%82%E8%AF%81%EF%BC%9A%7Ban%7D%E4%B9%9F%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97.)
7 设两个数列{an},{bn}满足bn=a1+a2+3a3+…nan/1+2+3…+n,若{bn}为等差数列,求证:{an}也为等差数列.
7 设两个数列{an},{bn}满足bn=a1+a2+3a3+…nan/1+2+3…+n,若{bn}为等差数列,求证:{an}也为等差数列.
7 设两个数列{an},{bn}满足bn=a1+a2+3a3+…nan/1+2+3…+n,若{bn}为等差数列,求证:{an}也为等差数列.
bn=a1+2a2+3a3+…nan/1+2+3…+n
b(n+1)=[a1+2a2+3a3+…nan+(n+1)a(n+1)]/[1+2+3…+n+(n+1)]
[n(n+1)/2]bn=a1+2a2+3a3+…nan ①
[(n+1)(n+2)/2]b(n+1)=a1+2a2+3a3+…nan+(n+1)a(n+1) ②
②-①得
[(n+1)(n+2)/2]b(n+1)-[n(n+1)/2]bn=(n+1)a(n+1)
两边同时消去(n+1)得
a(n+1)=[(n+2)/2]b(n+1)-(n/2)bn③
an=[(n+1)/2]bn-[(n-1)/2]b(n-1) ④
③-④得a(n+1)-an=[(n+1)/2]b(n+1)+1/2b(n+1)-[(n+1)/2]bn-[(n-1)/2]bn+[(n-1)/2]b(n-1)-1/2bn
=[(n+1)/2][b(n+1)-bn]+1/2[b(n+1)-bn]-[(n-1)/2][bn-b(n-1)]
又{bn}为等差数列,设公差为d
则a(n+1)-an=[(n+1)/2]d+1/2*d-[(n-1)/2]d
=3/2d
所以{an}是公差为3/2d的等差数列
注:此中的an,bn,a(n+1),b(n+1)均是数列中的项
看上去写的有点麻烦,但过程还是比较简单的