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Question Number 171189 by mathlove last updated on 09/Jun/22

$$f\left(x\right)=\frac{{16}^x}{4+{16}^x}faindvolueof$$$$f\left(\frac{1}{20}\right)+f\left(\frac{2}{20}\right)+........+f\left(\frac{19}{20}\right)$$

Answered by cortano1 last updated on 09/Jun/22

$$f\left(x\right)=\frac{{16}^x+4-4}{{16}^x+4}=1-\frac{4}{{16}^x+4}$$$$f\left(x\right)=1-\frac{1}{1+4^{2x-1}}$$$$f\left(\frac{1}{20}\right)=1-\frac{1}{1+4^{-\frac{18}{20}}}$$$$f\left(\frac{2}{20}\right)=1-\frac{1}{1+4^{-\frac{16}{20}}}$$$$f\left(\frac{3}{20}\right)=1-\frac{1}{1+4^{-\frac{14}{20}}}$$$$⋮$$$$f\left(\frac{19}{20}\right)=1-\frac{1}{1+4^{\frac{18}{20}}}$$$$\underset{k=1}{\sum ^{19}}f\left(\frac{k}{20}\right)=\underset{k=1}{\sum ^{19}}(1-\frac{1}{1+4^{\frac{2k-20}{20}}})$$$$=19-\underset{k=1}{\sum ^{19}}\left(\frac{1}{1+2^{\frac{k-10}{5}}}\right)$$$$=19-\frac{19}{2}=\frac{19}{2}$$$$$$

Commented by mathlove last updated on 09/Jun/22

$$thanks$$

Answered by Jamshidbek last updated on 09/Jun/22

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{4^{2\mathrm{x}-1}}{1+4^{2\mathrm{x}-1}}\mathrm{and}\mathrm{f}(1-\mathrm{x})=\frac{4^{-2\mathrm{x}+1}}{1+4^{-2\mathrm{x}+1}}=\frac{1}{4^{2\mathrm{x}-1}+1}$$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}(1-\mathrm{x})=1$$$$\mathrm{Use}\mathrm{this}\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}(1-\mathrm{x})=1$$

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