Question 351855
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Not.  Close but no cigar.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ \frac{n}{n\ +\ 1}]


Multiply both sides by *[tex \Large n\ +\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s(n\ +\ 1)\ =\ n]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ sn\ +\ s\ =\ n]


Add *[tex \Large -n] to both sides, and add *[tex \Large -s] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ sn\ -\ n\ =\ -s]


Factor *[tex \Large n] from the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n(s\ -\ 1)\ =\ -s]


Multiply both sides by *[tex \Large \frac{1}{s\ -\ 1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ =\ \frac{-s}{s\ -\ 1}]


See?  Your answer and the correct answer differ by a factor of -1.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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