Find the value of n that satisfies 2(n+1)!+6n!=3(n+1)!, where n! = n * (n-1) * (n-2)...2*1.
The answer is not 0.
Knowing that (n + 1)! = (n + 1)·n!:
2(n + 1)! + 6n! = 3(n + 1)!
2·(n + 1)·n! + 6·n! = 3·(n + 1)·n!
Divide both sides by n!:
2(n + 1) + 6 = 3(n + 1)
2n + 2 + 6 = 3n + 3
2n + 8 = 3n + 3
8 = n + 3
5 = n
+171 
