If $(n + 1)! = 12 (n - 1)!$ find the value of $n$ .

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Solution

$(n + 1)! = 12 (n - 1)!$
$(n (n + 1) - 12) (n - 1)! = 0$ $[n! = n (n - 1)!]$
As $(n - 1)!$ cannot be zero
$=> (n (n + 1) - 12) = 0$
$n^{2} + n - 12 = 0$

$n^{2} + 4 n - 3 n - 12 = 0$

$(n + 4) (n - 3) = 0$

Either $n = - 4$ or $n = 3$
For $n = - 4, (n - 1)!$ is not defined
$=>$ Value of n is 3

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