Find the value of n that satisfies 2*(n+1)!+6*n!=4*(n+1)!, where n! = n * (n-1) * (n-2)...2*1.
Hello Guest!
2*(n+1)!+6*n!=4*(n+1)!
6n! = 2(n+1)!
6n! = 2n! * (n+1)
3 = n + 1
2 = n
I hope this helped. :))
=^._.^=
