Question

已知f(x)=a1x+a2x^2+a3x^3+...+anx^n,且f(1)=n^2

Answer 1

依题可得：f(1)=a1+a2+a3+...+an=n^2

可知{an}是首项a1=1,公差为2的等差数列.

即an=2n-1 (n=1,2,3,...)

所以1/a1a2 1/a2a3 ...1/ana(n+1)=1/1*3 1/3*5 ...1/(2n-1)(2n+1)

= {(1-1/3)+(1/3-1/5)+...+[1/(2n-1)-1/(2n+1)]}*1/2

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

=n/(2n+1)

注：* 为乘号.