已知数列{an}中a1=1,2an=sn*s(n-1)求an

显示全部楼层 · 2012-10-13 12:25:37

即2[Sn-S(n-1)]=SnS(n-1)两边除以2SnS(n-1)1/S(n-1)-1/Sn=1/2所以1/Sn是等差数列，d=-1/2S1=a11/Sn=1/a1-1/2*(n-1)=(3-n)/2Sn=2/(3-n)所以n>=2an=Sn-S(n-1)=2/(3-n)-2/(4-n)=2/(n2-7n+12)所以an=1，n=12/(n2-7n+12)，n≥2...

千问 · 2012-10-13 12:25:37

2an=sn*s(n-1)2[sn-s(n-1)]=sn*s(n-1)1/s(n-1)-1/sn=1/21/sn-1/s(n-1)=-1/21/sn=1+(n-1)*(-1/2)=(3-n)/2sn=2/(3-n)an=sn-s(n-1)=2/(3-n)-2/(4-n)=2/(3-n)(4-n)(n≥2)an=1(n=1)...

千问 · 2012-10-13 12:25:37

2(Sn-S(n-1))=Sn*S(n-1)1/Sn-1/S(n-1)=-1/21/Sn=1-(n-1)/2=(3-n)/2Sn=2/(3-n)an=Sn-Sn-1=-2/(3-n)(4-n)a1=1...