Question

已知f(x)=a1x+a2x^2+…+anx^n,切对任意n属于正整数,都有f(1)=n^2 (1)求数列｛an｝的通项

Answer 1

an=n^2-(n-1)^2=2n-1;

f(1/2)=1/2+3/4+...+(2n-1)/2^n;

0.5*f(1/2)=1/4+3/8+…+ (2n-1)/2^(n+1);

两式子错位相减得0.5*f(1/2)=1/2+2/4+2/8+...+2/2^n-(2n-1)/2^(n+1);

f(1/2)=3/2-1/2^(n-1)-(2n-1)/2^(n+1);