Question

已知数列an的前n项和sn满足：sn=a-1分之a（an-1）（a为常数,且a不等于0,a不等于1）求an的通项公式

Answer 1

1)当n=1时,S1=a1=a/(a-1)(a1-1),a1=ax0d当n≥2时,an=Sn-S(n-1)=a/（a-1)（an-a(n-1)）移项得x0dan=a*a(n-1),即an/a(n-1)=ax0d因此{an}是公比为a的等比数列x0d所以{an}的通项公式为an=a^n2)由1）可得Sn=a(a^n-1)/(a-1)x0d于是bn=（2Sn）/an 1=2a(a^n-1)/(a^n(a-1)) 1x0d=2a/(a-1)-2a/(a^n(a-1)) 1x0d若{bn}为等比数列,则由bn/b(n-1)=b(n 1)/bn