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Find n if, ...

Question

Find $n$ if, $(n + 1)! = 42 (n - 1)!$

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Solution

$(n + 1)! = 42 (n - 1)!$
$\Rightarrow (n + 1) (n) (n - 1)! = 42 (n - 1)$
$\Rightarrow n^{2} + n - 42 = 0$
$\Rightarrow n^{2} + 7 n - 6 n - 42 = 0$
$\Rightarrow n (n + 7) - 6 (n + 7) = 0$
$\Rightarrow (n + 7) (n - 6) = 0$
$\Rightarrow n = - 7, n = 6$
$∴ n = 6$ .

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