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Find the value of n that satisfies n-2(n+1)!+6n!=3(n+1), where n!=n*(n-1)*(n-2)***2*1 .

ANotSmartPerson Oct 24, 2018

#1**+1 **

I don't think the LHS of your equation will ever equal the RHS as it is written. Can you see why?

Guest Oct 25, 2018

#3**+1 **

Lets see

\(n-2(n+1)!+6n!=3(n+1)\\ n-2n!(n+1)+6n!=3n+3\\ -2n!(n+1)+6n!=2n+3\\ n![-2(n+1)+6]=2n+3\\ n![-2n-2+6]=2n+3\\ n![4-2n]=2n+3\\ n!=\frac{2n+3}{-2n+4}\\ \)

if n=1 LHS=1 RHS=5/2 = 2.5

IF N=2 LHS=2 RHS=7/0 UNDEFINED

IF N=3 LHS=6 RHS=9/-2 = -4.5

For all higher integer values of n the right hand side will be negative so there are no solutions to this.

(unless i made some stupid algebraic error)

Melody Oct 26, 2018