Problem Solving (PS) | Re: If n* = 1 - 1/(1 - n) for any none-zero integer n, then, n* (n 1)*

Given that\( n^* = 1 -\frac{ 1}{1 -n}\) and we need to find the value of\( n^* – (n –1)^*\)

\( n^* = 1 -\frac{ 1}{1 -n}\) =\( 1 - n - 1/1 -n\) =\(\frac{-n}{1-n}\) =\(\frac{ n}{n-1}\)


=> to find\( (n –1)^*\) we need to replace n with n-1 in\( n^* =\frac{ n}{n-1}\)
=>\( (n-1)^* =\frac{ n-1}{ n-1-1}\) =\(\frac{ n - 1}{ n -2}\)

=>\( n^* – (n –1)^*\) =\(\frac{ n}{n-1}\) -\(\frac{ n - 1}{ n -2}\)
=\(\frac{ n*(n -2) - (n-1)^2}{ (n - 1)*(n -2)}\) =\(\frac{ n^2 -2n - n^2 + 2n - 1}{ (n - 1)*(n -2)}\)
=\(\frac{ -1}{ (n - 1)*(n -2)}\) =\(\frac{ 1}{(1 - n)(n -2)}\)

So, Answer will beA
Hope ithelps!

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Statistics : Posted by BrushMyQuant • on 06 Sep 2022, 12:15 • Replies 2 • Views 765

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