Question

3((n+3)!-(n+1)!)=2((n+3)!+(n+1)!) calculate n value ?

3((n+3)!-(n+1)!)=2((n+3)!+(n+1)!) calculate n value ?

Answer 1

`n=(-5+-sqrt(25-4))/2`

Explanation:

We have `3((n+3)!-(n+1)! =2((n+3)!+(n+1)!)`

or `3((n+3)xx(n+2)xx(n+1)!-(n+1)!) = 2((n+3)xx(n+2)xx(n+1)!+(n+1)!)`

or `3(n+1)![(n+3)(n+2)-1]=2(n+1)![(n+3)(n+2)+1]`

or `3[(n+3)(n+2)-1])=2[(n+3)(n+2)+1]`

or `3(n^2+5n+5)=2(n^2+5n+7)`

or `n^2+5n+1=0`

i.e. `n=(-5+-sqrt(25-4))/2`

Answer 2

The solution is `n in {(-5+sqrt21)/2, (-5-sqrt21)/2}`

Explanation:

The equation is

`3((n+3)!-(n+1)!)=2((n+3)!+(n+1)!)`

Factor `(n+1)!`

`3(n+1)!((n+3)(n+2)-1)=2(n+1)!((n+3)(n+2)+1)`

Simplifying by `(n+1)!`

`3((n+3)(n+2)-1)=2((n+3)(n+2)+1)`

`3(n^2+5n+6-1)=2(n^2+5n+6+1)`

`3n^2+15n+15=2n^2+10n+14`

`n^2+5n+1=0`

This is a quadratic equation in `n`

The discriminant is

`Delta=5^2-4=25-4=21`

`n=(-5+-sqrt21)/(2)`

There is a contradiction since the definition of factorials `n in NN`

And here

`n in RR`

Answer 3

`n=(-5+-sqrt(21))/(2)`

Explanation:

`3((n+3)!-(n+1)!)=2((n+3)!+(n+1)!)`
Collecting like terms
`3(n+3)!-3(n+1)! =2(n+3)!+2(n+1)!`
`3(n+3)!-2(n+3)! =2(n+1)!+3(n+1)!`
`(n+3)! =5(n+1)!`

Using Property (`(n! =n(n-1)! =n(n-1)(n-2)!)`

`(n+3)(n+2)cancel((n+1)!) =5cancel((n+1)!)`
`n^2+5n+6=5`
`n^2+5n+1=0`
Using Quadratic Solution `x=(-b+-sqrt(b^2-4ac))/(2a)`
`a=1, b=5, c=1`
`n=(-5+-sqrt(5^2-4(1)(1)))/(2(1))`
`n=(-5+-sqrt(25-4))/(2)`
`n=(-5+-sqrt(21))/(2)`