已知数列{an}的首项a1不等于0,其前n项和为Sn,且S(n+1)=2Sn+a1,求lim(an/Sn)的值,
答案解析
举报
S (n+1)=2S n + a1 .(1)
S n =2S (n-1) +a1.(2)
(1)-(2)得
S (n+1)-Sn=2[S n -S (n-1)]
a (n+1)=2 an
∴an是q=2的等比数列
an=a1X2^(n-1)
Sn=a1(1-2^n)/(1-2)=a1(2^n-1)
liman/Sn=lim[2^(n-1)/(2^n-1)]=1/2
解析看不懂?求助智能家教解答
查看解答
更多答案(3)