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

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

$$S=\frac{1}{1+e^{-99}}+\frac{1}{1+e^{-98}}+\cdot \cdot \cdot +\frac{1}{1+e^{98}}+\frac{1}{1+e^{99}}$$$$=\frac{e^{99}}{1+e^{99}}+\frac{e^{98}}{1+e^{98}}+\cdot \cdot \cdot +\frac{1}{2}+\cdot \cdot \cdot +\frac{1}{1+e^{98}}+\frac{1}{1+e^{99}}$$$$=\frac{1+e^{99}}{1+e^{99}}+\frac{1+e^{98}}{1+e^{98}}+\cdot \cdot \cdot +\frac{1}{2}=99+\frac{1}{2}=\frac{199}{2}$$

Commented by mathlove last updated on 31/Jan/22

$$thanks$$

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

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

Commented by mathlove last updated on 31/Jan/22

$$thanks$$

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