evdo wrote:

\(\frac{(n+1)}{(1-n)}=-\frac{(n+1)}{(n-1)}\)

How do you do this kind of operation? Can you share any study materials on this topic? Thanks

This is basic algebra and fraction.

You take our "-" common from the denominator to get

\(\frac{(n+1)}{(1-n)}=\frac{(n+1)}{-(n-1)}\)

And from theory of fractions, you know that

\(\frac{(n+1)}{-(n-1)}=-\frac{(n+1)}{(n-1)} = \frac{-(n+1)}{(n-1)}\)

For these basics, check purplemath.com or khanacademy.org

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