Question

已知数列{an}和{bn},{an}的前n项和为Sn,a2=0,且对任意n∈N*都有2Sn=n(an-1),点列Pn（a

Answer 1

2*a1=2*a1=a1-1

a1=-1

n>=2时

2*Sn=n*an-n

2*S(n-1)=n*a(n-1)-a(n-1)-n+1

(n-2)an=(n-1)a(n-1)+1

n>=3时

an/(n-1)=a(n-1)/(n-2)+1/((n-2)(n-1))

a2=0

an/(n-1)=a2+1/(1*2)+1/(2*3)+……+1/((n-2)(n-1))=(n-2)/(n-1)

an=n-2

代入n=1,2时亦符合

故an=n-2

bn=2*an+2=2n-2

Pn(n-2,2n-2)

P1(-1,0)

(P1Pn)^2=(n-2+1)^2+(2n-2)^2=5*(n-1)^2

原式=(1/5)*(1/1^2+1/2^2+1/3^2+……+1/(n-1)^2)