+0

0
109
3
+176

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
+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
+176
0

Someone else helped me, the answer is 5.

Oct 26, 2018
edited by ANotSmartPerson  Oct 26, 2018
#3
+95017
+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