Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 165364 by mathlove last updated on 31/Jan/22

Answered by Ar Brandon last updated on 31/Jan/22

S=(1/(1+e^(−99) ))+(1/(1+e^(−98) ))+∙∙∙+(1/(1+e^(98) ))+(1/(1+e^(99) ))      =(e^(99) /(1+e^(99) ))+(e^(98) /(1+e^(98) ))+∙∙∙+(1/2)+∙∙∙+(1/(1+e^(98) ))+(1/(1+e^(99) ))      =((1+e^(99) )/(1+e^(99) ))+((1+e^(98) )/(1+e^(98) ))+∙∙∙+(1/2)=99+(1/2)=((199)/2)

$${S}=\frac{\mathrm{1}}{\mathrm{1}+{e}^{−\mathrm{99}} }+\frac{\mathrm{1}}{\mathrm{1}+{e}^{−\mathrm{98}} }+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{98}} }+\frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{99}} } \\ $$$$\:\:\:\:=\frac{{e}^{\mathrm{99}} }{\mathrm{1}+{e}^{\mathrm{99}} }+\frac{{e}^{\mathrm{98}} }{\mathrm{1}+{e}^{\mathrm{98}} }+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{2}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{98}} }+\frac{\mathrm{1}}{\mathrm{1}+{e}^{\mathrm{99}} } \\ $$$$\:\:\:\:=\frac{\mathrm{1}+{e}^{\mathrm{99}} }{\mathrm{1}+{e}^{\mathrm{99}} }+\frac{\mathrm{1}+{e}^{\mathrm{98}} }{\mathrm{1}+{e}^{\mathrm{98}} }+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{99}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{199}}{\mathrm{2}} \\ $$

Commented by mathlove last updated on 31/Jan/22

thanks

$${thanks} \\ $$

Answered by Ar Brandon last updated on 31/Jan/22

S=Σ_(n=1) ^(99) ((1/(1+e^(−n) ))+(1/(1+e^n )))+(1/2)     =Σ_(n=1) ^(99) (((e^n +1)/(e^n +1)))+(1/2)=99+(1/2)

$${S}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{1}+{e}^{−{n}} }+\frac{\mathrm{1}}{\mathrm{1}+{e}^{{n}} }\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\left(\frac{{e}^{{n}} +\mathrm{1}}{{e}^{{n}} +\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{99}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by mathlove last updated on 31/Jan/22

thanks

$${thanks} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com