Question

已知数列｛an｝的前n项和Sn满足：Sn=a/a-1（an-1)(a为常数,且a不等于0,a不等于1）.求数列｛an｝的

Answer 1

答：n=1时,a[1]=s[1]=a/(a-1)(a[1]-1),a[1]=a;

a[n+1]=s[n+1]-s[n]=a/(a-1)*(a[n+1]-a[n]),解得：a[n+1]=aa[n],故a[n]=a[1]a^(n-1)=a^n

(2)将a[n]=a^n代入b[n]=（2S[n]/a[n]）+1中得：b[n]=(3a-2a^(1-n)-1)/(a-1),若数列｛bn｝为等比数列,

则b[1],b[2],b[3]必成等比列,由此可推出：a=1/3;

(3)c[n]=1/{1+(1/3)^n}+1/{1-(1/3)^(n+1)}=3^n/(1+3^n)+3^(n+1)/(3^(n+1)-1)=2+1/{3^(n+1)-1}-1/(1+3^n)>2-{1/3^n-1/3^(n+1)}

故T[n]>2n-{1/3-1/3^(n+1)}>2n-1/3