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Question Number 158306 by mathlove last updated on 02/Nov/21

Commented by cortano last updated on 02/Nov/21

$$f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}=\frac{\frac{4}{4^x}}{\frac{4}{4^x}+2}=\frac{4}{4+2.4^x}$$$$f\left(x\right)+f(1-x)=\frac{4^x}{4^x+2}+\frac{4}{2.4^x+4}$$$$=\frac{2.4^x+4}{2.4^x+4}=1$$$$\Rightarrow f\left(\frac{1}{1997}\right)+f\left(\frac{2}{1997}\right)+...+f\left(\frac{1996}{1997}\right)=\frac{1998}{2}=999$$

Commented by mathlove last updated on 02/Nov/21

$$whyfaindf(1-x)$$$$andwhytheend\frac{1998}{2}$$

Commented by tounghoungko last updated on 03/Nov/21

$$since\frac{1996}{1997}=1-\frac{1}{1997}$$$$\frac{1995}{1997}=1-\frac{2}{1997}.andsosoon$$

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