Question

已知数列{an}的前n项和为Sn,且曲线y=x^2-nx+1(n∈N)在x=an处的切线的斜率恰好为Sn 1求数列{an

Answer 1

(1)

y=x^2-nx+1

y'=2x-n

y'(an)=2*an-n=Sn

a1=S1=2*a1-1

a1=1

Sn=2*an-n

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

an-2*a(n-1)-1=0

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

a1+1=2

an=2^n-1

(2)

n*an=n*2^n-n

Tn=n*2^(n+1)-(2^1+2^2+…+2^n)-n(n+1)/2=n*2^(n+1)-2^(n+1)+2-n(n+1)/2

(3)

1/a1+1/a2+1/a3+……+1/an=1/(2^1-1)+1/(2^2-1)+…+1/(2^n-1)