记数列{an}的前n项和为Sn,若{Sn/an}是公差为d的等差数列,则{an}为等差数列时d=解释的很详尽 但是当d≠1时,an-1=(n-1)(1-d)•d′/d,这一步看不懂...想知道是怎么得到的∵{Sn/an}是S1a1=1为首项,d
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![记数列{an}的前n项和为Sn,若{Sn/an}是公差为d的等差数列,则{an}为等差数列时d=解释的很详尽 但是当d≠1时,an-1=(n-1)(1-d)•d′/d,这一步看不懂...想知道是怎么得到的∵{Sn/an}是S1a1=1为首项,d](/uploads/image/z/13636229-5-9.jpg?t=%E8%AE%B0%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E8%8B%A5%7BSn%2Fan%7D%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%BAd%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%88%99%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E6%97%B6d%3D%E8%A7%A3%E9%87%8A%E7%9A%84%E5%BE%88%E8%AF%A6%E5%B0%BD+%E4%BD%86%E6%98%AF%E5%BD%93d%E2%89%A01%E6%97%B6%2Can-1%3D%EF%BC%88n-1%EF%BC%89%EF%BC%881-d%EF%BC%89%26%238226%3Bd%E2%80%B2%2Fd%2C%E8%BF%99%E4%B8%80%E6%AD%A5%E7%9C%8B%E4%B8%8D%E6%87%82...%E6%83%B3%E7%9F%A5%E9%81%93%E6%98%AF%E6%80%8E%E4%B9%88%E5%BE%97%E5%88%B0%E7%9A%84%E2%88%B5%7BSn%2Fan%7D%E6%98%AFS1a1%3D1%E4%B8%BA%E9%A6%96%E9%A1%B9%2Cd)
记数列{an}的前n项和为Sn,若{Sn/an}是公差为d的等差数列,则{an}为等差数列时d=解释的很详尽 但是当d≠1时,an-1=(n-1)(1-d)•d′/d,这一步看不懂...想知道是怎么得到的∵{Sn/an}是S1a1=1为首项,d
记数列{an}的前n项和为Sn,若{Sn/an}是公差为d的等差数列,则{an}为等差数列时d=
解释的很详尽 但是
当d≠1时,an-1=(n-1)(1-d)•d′/d,这一步看不懂...想知道是怎么得到的
∵{Sn/an}是S1a1=1为首项,d为公差的等差数列,
∴Snan=1+(n-1)d,
∴Sn=an+(n-1)dan,①
Sn-1=an-1+(n-2)dan-1.②
①-②得:
an=an+(n-1)dan-an-1-(n-2)dan-1,
整理可得
(n-1)dan-(n-1)dan-1=(1-d)an-1,
假设d=0,那么Sn/an=1,
S1=a1,S2=a1+a2=a2,
∴a1=0,∵a1为除数,不能为0,∴d≠0.
在此假设an的公差为d′,
所以有d′=(1-d)an-1(n-1)d,
当d=1时,d′=0,an是以a1为首项,0为公差的等差数列.
当d≠1时,an-1=(n-1)(1-d)•d′/d,
an-an-1=(1-d)•d′/d=d′,
∴d=1/2,
此时,an是以d′为首项,d′为公差的等差数列.
综上所述,d=1,或d=1/2.
1或1/2.
有一个步骤看不懂
当d≠1时,an-1=(n-1)(1-d)•d′/d,
这一步是怎么得到的?
记数列{an}的前n项和为Sn,若{Sn/an}是公差为d的等差数列,则{an}为等差数列时d=解释的很详尽 但是当d≠1时,an-1=(n-1)(1-d)•d′/d,这一步看不懂...想知道是怎么得到的∵{Sn/an}是S1a1=1为首项,d
根据等差数列递增规律
易证得
da2+(d+1)/2d a2=2a2
所以d=1 / 1/2