Question

暑假作业数学题，想了很久都不会，求大神解答，感激不尽！

Answer 1

①将an=Sn-Sn-1代入Sn^2=an(Sn-1/2)(n>2)

可得Sn^2=(Sn-Sn-1)(Sn-1/2)=Sn^2-1/2×Sn-SnSn-1+1/2×Sn-1

合并同类项可得-Sn-2SnSn-1+Sn-1=0两边除以SnSn-1

得到1/Sn-1/Sn-1=2

即1/Sn是以1/S1=1/a1=1为首项，2为公差的等差数列

所以求得1/Sn=1+2(n-1)=2n-1

因此Sn=1/2n-1

而Sn-1=1/2n-3

an=Sn-Sn-1=1/2n-1-1/2n-3=-2/(2n-1)(2n-3)

②bn=Sn/2n+1见附件

追问

大神再问你两题行吗