If (n+1)! = 6[(n-1)!],find n. | vedclass

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(D) Given the equation: $(n+1)! = 6(n-1)!$We know that $n! = n 	imes (n-1)!$,so $(n+1)! = (n+1) 	imes n 	imes (n-1)!$.Substituting this into the eq

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$2$

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(D) Given the equation: $(n+1)! = 6(n-1)!$We know that $n! = n 	imes (n-1)!$,so $(n+1)! = (n+1) 	imes n 	imes (n-1)!$.Substituting this into the equation:$(n+1) 	imes n 	imes (n-1)! = 6(n-1)!$Since $(n-1)!$ is non-zero,we can divide both sides by $(n-1)!$:$(n+1) 	imes n = 6$$n^2 + n - 6 = 0$Factoring the quadratic equation:$(n+3)(n-2) = 0$This gives $n = -3$ or $n = 2$.Since $n$ must be a natural number for the factorial to be defined,we discard $n = -3$.Therefore,$n = 2$.

If $(n+1)! = 6[(n-1)!]$,find $n$.

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