If $F (n) = (- 1)^{k - 1} (n - 1), G (n) = n - F (n)$ then $(G o G) (n) =$ (where $k$ is odd)

1

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n

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2

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n - 1

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Solution

The correct option is B $1$
$G (n) = n - (- 1)^{k - 1} (n - 1)$
$G o G (n) = G (n - (- 1)^{k - 1} (n - 1))$
$= n - (- 1)^{k - 1} (n - 1) - (- 1)^{k - 1} ((n - 1) - (- 1)^{k - 1} (n - 1))$
$= n - (n - 1)$
$= 1$

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