Simplify the following and express with positive index :
[1 - {1 - (1 - n)-1}-1]-1
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Simplifying the expression :
⇒[1−{1−(1−n)−1}−1]−1=[1−{1−11−n}−1]−1=[1−{1−n−11−n}−1]−1=[1−{−n1−n}−1]−1=[1−{−1−nn}]−1=[1+1−nn]−1=[n+1−nn]−1=[1n]−1=n.\Rightarrow [1 - {1 - (1 - n)^{-1}}^{-1}]^{-1} = \Big[1 - \Big{1 - \dfrac{1}{1 - n}\Big}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \Big{\dfrac{1 - n - 1}{1 - n}\Big}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \Big{\dfrac{-n}{1 - n}\Big}^{-1}\Big]^{-1} \\[1em] = \Big[1 - \Big{-\dfrac{1 - n}{n}\Big}\Big]^{-1} \\[1em] = \Big[1 + \dfrac{1 - n}{n}\Big]^{-1} \\[1em] = \Big[\dfrac{n + 1 - n}{n}\Big]^{-1} \\[1em] = \Big[\dfrac{1}{n}\Big]^{-1}\\[1em] = n.⇒[1−{1−(1−n)−1}−1]−1=[1−{1−1−n1}−1]−1=[1−{1−n1−n−1}−1]−1=[1−{1−n−n}−1]−1=[1−{−n1−n}]−1=[1+n1−n]−1=[nn+1−n]−1=[n1]−1=n.
Hence, [1−{1−(1−n)−1}−1]−1=n[1 - {1 - (1 - n)^{-1}}^{-1}]^{-1} = n[1−{1−(1−n)−1}−1]−1=n.
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