Question

已知函数f(x)=a1x a2x^2 a3x^3 … anx^n,且对一切正整数n都有f(1)=n^2成立

Answer 1

1,设Sn=a1+a2+a3+…+an S1=a1=1^2=1

n>=2时,an=Sn-S(n-1)=n^2-(n-1)^2=2n-1 a1=1也符合此式.

所以通项公式为：an=2n-1(n=1,2,3,…)

2,1/(a1a2)+1/(a2a3)+…+1/[a(n-1)an]

=1/(1*3)+1/(3*5)+…+1/[(2n-3)(2n-1)]

=(1/2)[1-1/3+1/3-1/5+…+1/(2n-3)-1/(2n-1)]

=(1/2)[1-1/(2n-1)]

=(n-1)/(2n-1)