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Question Number 83542 by Power last updated on 03/Mar/20

Answered by john santu last updated on 03/Mar/20

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{1+2^{2\mathrm{x}-1}}$$$$\mathrm{f}\left(\frac{1}{2001}\right)=\frac{1}{1+2^{\frac{2}{2001}-1}}$$$$\mathrm{f}\left(\frac{2}{2001}\right)=\frac{1}{1+2^{\frac{2}{2001}-1}}$$$$\mathrm{f}\left(\frac{3}{2001}\right)=\frac{1}{1+2^{\frac{3}{2001}-1}}$$$$\mathrm{f}\left(\frac{1}{2001}\right)+\mathrm{f}\left(\frac{2}{2001}\right)+\mathrm{f}\left(\frac{3}{2001}\right)+...+$$$$\mathrm{f}\left(\frac{2000}{2001}\right)=\frac{1}{1+2^{\frac{1}{2001}-1}}+\frac{1}{1+2^{\frac{2}{2001}-1}}+$$$$\frac{1}{1+2^{\frac{3}{2001}-1}}+...+\frac{1}{1+2^{\frac{2000}{2001}-1}}=$$

Commented by john santu last updated on 03/Mar/20

$$=\frac{1}{1+2^{-\frac{2000}{2001}}}+\frac{1}{1+2^{-\frac{1999}{2001}}}+\frac{1}{1+2^{-\frac{1998}{2001}}}+$$$$...+\frac{1}{1+2^{-\frac{1}{2001}}}$$

Answered by mahdi last updated on 03/Mar/20

$$\mathrm{if}\mathrm{m}+\mathrm{n}=1\Rightarrow$$$$\mathrm{f}\left(\mathrm{m}\right)+\mathrm{f}\left(\mathrm{n}\right)=\frac{2}{4^{\mathrm{m}}+2}+\frac{2}{4^{\mathrm{n}}+2}=\frac{2(4^{\mathrm{n}}+2)+2(4^{\mathrm{m}}+2)}{(4^{\mathrm{m}}+2)(4^{\mathrm{n}}+2)}=$$$$\frac{8+2(4^{\mathrm{m}}+4^{\mathrm{n}})}{4+4^{\mathrm{m}}\times 4^{\mathrm{n}}+2(4^{\mathrm{m}}+4^{\mathrm{n}})}=1(4^{\mathrm{m}}\times 4^{\mathrm{n}}=4^{\mathrm{m}+\mathrm{n}}=4^1)$$$$\Rightarrow [\mathrm{f}\left(\frac{1}{2001}\right)+\mathrm{f}\left(\frac{2000}{2001}\right)]+[\mathrm{f}\left(\frac{2}{2001}\right)+\mathrm{f}\left(\frac{1999}{2001}\right)]+$$$$...+[\mathrm{f}\left(\frac{1000}{2001}\right)+\mathrm{f}\left(\frac{1001}{2001}\right)]=1000\times 1=1000$$

Commented by Power last updated on 03/Mar/20

$$\mathrm{thank}\mathrm{you}$$

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