Question

已知f(x)=a1x+a2x^2+...+anx^n(n∈N*),f(1)=n^2,求f(1/3)的值。

我算的是1-n/3^n...可是答案是1-n+1/3^n...麻烦各位帮忙算下了~谢谢~

Answer 1

f(1)=a1+a2+a3+....an=n^2
San=n^2
an=San-San-1=n^2-(n-1)^2=2n-1
f(x)=x+3x^2+.....(2n-1)x^n
f(1/3)=1/3+3*(1/3)^2+....(2n-1)(1/3)^n
1/3f(1/3)=(1/3)^2+3(1/3)^3+.....(2n-3)(1/3)^n+(2n-1)(1/3)^(n+1)
f(1/3)-1/3f(1/3)=1/3+2(1/3)^2+2(1/3)^3+.....2(1/3)^n-(2n-1)(1/3)^(n+1)
=1/3+2[(1/3)^2-(1/3)^(n+1)]/(1-1/3)-(2n-1)(1/3)^(n+1)
=1/3+1/3-(1/3)^n-(2n-1)(1/3)^(n+1)
=2/3-(1/3)^n-(2n-1)(1/3)^(n+1)
=2/3-(2n+2)(1/3)^(n+1)
即 2/3f(1/3)=2/3-(2n+2)(1/3)^(n+1)
f(1/3)=1-(n+1)/3^n