已知數列{an}的前n項和Sn,且滿足an+Sn×S n-1 =0(n≥2),a1=1/2.(1)求證{1/Sn}是等差數列(2)求an的表達式

熱心網友

(1):因為n≥2時Sn-S(n-1)=an,所以Sn-S(n-1)+Sn×S(n-1)=0即S(n-1)-S(n)=Sn×S(n-1),兩邊同時除以Sn×S(n-1)得:1/S(n)-1/S(n-1)=1,所以數列{1/S(n)}是一個以1/S(1)為首項,以1為公差的等差數列(2):由(1)得1/S(n)=1/S(1)+(n-1)=2+n-1=n+1,所以S(n)=1/(n+1)所以n≥2時:S(n-1)=1/n,所以an=Sn-S(n-1)=1/(n+1)-1/n=-1/n(n+1),當n=1時:an=1/2

熱心網友

(1)an+Sn×S n-1 =0==0=1/an+1/[Sn×S (n-1)] =1/an+1/[(S(n-1)+an)×S (n-1)] ==1/an-1/{an[(S(n-1)+an)]}+1/{an[S (n-1)] }==1/Sn=1/[S (n-1)] +1(2)1/Sn=1/[S (n-1)] +1=1/[S (n-2)] +2=..=1/[S1] +n-1=n+1==Sn=1/(n+1),an=Sn-S(n-1)=-1/[n(n+1)],n1.