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 ?

3 Answers
Oct 1, 2017

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

Oct 1, 2017

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

Oct 1, 2017

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)