Question

等差数列问题

1.在数列{an}中，相邻两项an和an+1是相应的二次方程x平方+3nx+bn=0（n∈N*）的两根，若a1=2，试求b100的值。
2.设数列{an}中a1=1且an=2Sn平方/2Sn-1（n≥2），证明{1/Sn}是等差数列

Answer 1

(an)+(an+1)=-3n (an+1)+(an+2)=-3n-3
得(an+2)-(an)=-3
因为a1=2所以a101=2+50*(-3)=-148
因为a2=-5所以a100=-5+49*(-3)=-152
得b100=a100*a101=22496

1/Sn-1/S(n-1)=-an/(S(n-1)*Sn)=-2Sn/(2Sn-1)(Sn-an)=-2/(2Sn-1)(1-2Sn/(2Sn-1))=2

Answer 2

l don't know.