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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 .

 Oct 24, 2018
 #1
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+1

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

 Oct 25, 2018
 #2
avatar+183 
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Someone else helped me, the answer is 5.

 Oct 26, 2018
edited by ANotSmartPerson  Oct 26, 2018
 #3
avatar+99122 
+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)

 Oct 26, 2018

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