由是=1为首项,d为公差的等差数列,知Sn=an+(n-1)dan,故Sn-1=an-1+(n-2)dan-1.所以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,由此入手,能够求出d.
【解析】
∵是=1为首项,d为公差的等差数列,
∴=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,
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′=,
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),
an-an-1==d′,
∴d=,
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=.
故答案为:1或.