Find the value of n that satisfies 2(n+1)!+6n!=4(n+1)!, where n! = n\cdot (n-1)\cdot (n-2) \cdots 2\cdot 1.
+1366 We have the equation: \(2(n+1)!+6n!=4(n+1)!\)
We can subtract \(2(n+1)!\) from both sides, which yields: \(6n!=2(n+1)!\)
Note that \((n+1)! = n! \times (n+1)\)
Substituting this gives us: \(6n!=2(n+1)n!\)
Now, we just have to divide by \(n!\), then solve for n.
Can you take it from here?
