If (n+2)! = 2550(n!),find n. | vedclass

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(C) Given the equation: $(n+2)! = 2550(n!)$We know that $(n+2)! = (n+2)(n+1)(n!)$.Substituting this into the equation:$(n+2)(n+1)(n!) = 2550(n!)$Since

Vedclass · Accepted Answer

$49$

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(C) Given the equation: $(n+2)! = 2550(n!)$We know that $(n+2)! = (n+2)(n+1)(n!)$.Substituting this into the equation:$(n+2)(n+1)(n!) = 2550(n!)$Since $n! 
eq 0$,we can divide both sides by $n!$:$(n+2)(n+1) = 2550$Expanding the left side:$n^2 + 3n + 2 = 2550$Rearranging into a quadratic equation:$n^2 + 3n - 2548 = 0$Factoring the quadratic equation:$n^2 + 52n - 49n - 2548 = 0$$n(n + 52) - 49(n + 52) = 0$$(n - 49)(n + 52) = 0$This gives two possible values for $n$:$n = 49$ or $n = -52$.Since $n$ must be a natural number $(n \in \mathbb{N})$,we reject $n = -52$.Therefore,$n = 49$.

If $(n+2)! = 2550(n!)$,find $n$.

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