We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# help please

0
196
3

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

### 3+0 Answers

#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
0

Someone else helped me, the answer is 5.

Oct 26, 2018
edited by ANotSmartPerson  Oct 26, 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)

Oct 26, 2018