Question

Cho \(f\left(x\right)=\frac{4^x}{4^x+2}\) 

tính \(S=f\left(\frac{1}{2007}\right)+f\left(\frac{2}{2007}\right)+........+f\left(\frac{2006}{2007}\right)\)

Answer 1

Ta có nhận xét : \(a+b=1\) thì

\(f\left(a\right)+f\left(b\right)=\frac{4^a}{4^a+2}+\frac{4^b}{4^b+2}=\frac{4^a\left(4^a+2\right)+4^b\left(4^b+2\right)}{\left(4^a+2\right)\left(4^b+2\right)}\)

                 \(=\frac{4^{a+b}+2.4^a+4^{a+b}+2.4^b}{4^{a+b}+2.4^a+2.4^b+4}=\frac{2.4^a+2.4^b+8}{2.4^a+2.4^b+8}=1\)

Áp dụng kết quả trên ta có :

\(S=\left[f\left(\frac{1}{2007}\right)+f\left(\frac{2006}{2007}\right)\right]+\left[f\left(\frac{2}{2007}\right)+f\left(\frac{2005}{2007}\right)\right]+...+\left[f\left(\frac{1003}{2007}\right)+f\left(\frac{1004}{2007}\right)\right]\)

Vâyu \(S=1+1+1+...+1=1003\) (có 1003 số hạng)