Find the value of $n$ that satisfies $2(n+1)!+6n!=3(n+1)!$, where $n! = n\cdot (n-1)\cdot (n-2) \cdots 2\cdot 1$.
2 ( n+ 1) ! + 6n! = 3 ( n + 1)!
Note that ( n + 1)! = (n + 1) * n!
So
2 (n + 1) * n! + 6n! = 3 (n + 1) * n! divide out n!
2 ( n + 1) + 6 = 3 ( n + 1) simplify
2n + 2 + 6 = 3n + 3
8 = n + 3
8 - 3 = n = 5
Thanks Chris
