a1=1,n≥2时 ,sn2=an[sn-(1/2)]
问题描述:
a1=1,n≥2时 ,sn2=an[sn-(1/2)]
求sn
答
由题意:(a1+a2)^2=a2*[a1+a2-1/2)] 即(1+a2)^2=a2(a2+1/2)解得:a2=-2/3 n>=2时,因为(Sn)^2=an*(Sn -1/2)故(Sn)^2=(Sn-Sn-1) (Sn -1/2) 化简得到:Sn-1Sn=1/2* Sn-1 -1/2*Sn 即1/Sn -1/Sn-1 =2故故数列{1/S...