Factorial

n!=1·2·3…·n
  = (n-1)! · n

0! = 1

Coeficiente Binomial

Si 0≤k≤n el coeficiente binomial

⎛n⎞      n!
⎝k⎠ = __________
       k!(n-k)!

To provide an inductive definition of $( n! )$ (factorial of $( n )$ ), we start by defining the factorial function for a base case and then define it for any positive integer $( n )$ in terms of the factorial of $( n - 1 )$.

Inductive Definition of Factorial

1. Base Case: Define $( 0! )$ as follows:

   0! = 1.
   

2. Inductive Step: For any integer $( n \geq 1 )$, define $( n! )$ in terms of $( (n - 1)! )$ as:

   n! = n \cdot (n - 1)!.
   

This definition works because each factorial value is expressed recursively, relying on the factorial of the previous integer.

Summary of the Inductive Definition

• Base Case: $( 0! = 1 )$.
• Inductive Step: For $( n \geq 1 )$, $( n! = n \cdot (n - 1)! )$.

Explanation

Using this inductive definition, we can compute factorial values step-by-step:

• $( 1! = 1 \cdot 0! = 1 \cdot 1 = 1 )$,
• $( 2! = 2 \cdot 1! = 2 \cdot 1 = 2 )$,
• $( 3! = 3 \cdot 2! = 3 \cdot 2 = 6 )$,
• and so on.

This inductive approach to defining factorials is widely used in mathematical definitions and recursive programming.

To prove that $( 2^n < n! )$ for all $( n \geq 4 )$, we’ll use mathematical induction.

Step 1: Base Case

We start by verifying the inequality for $( n = 4 )$.

1. Calculate $( 2^4 )$:

   2^4 = 16.
   

2. Calculate $( 4! )$:

   4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24.
   

Since $( 16 < 24 )$, the inequality $( 2^4 < 4! )$ holds. Thus, the base case is true.

Step 2: Inductive Hypothesis

Assume that the inequality holds for some $( n = k \geq 4 )$. That is, assume:

2^k < k!.

This assumption is called the inductive hypothesis.

Step 3: Inductive Step

We need to prove that if the inequality holds for $( n = k )$, then it also holds for $( n = k + 1 )$. In other words, we need to show:

2^{k+1} < (k + 1)!.

Proof of the Inductive Step

Starting from $( 2^{k+1} )$, we can express it as:

2^{k+1} = 2 \cdot 2^k.

By the inductive hypothesis, we know that $( 2^k < k! )$. Substituting this inequality into the expression above, we get:

2^{k+1} < 2 \cdot k!.

Now, we need to show that:

2 \cdot k! \leq (k + 1)!.

Since $( (k + 1)! = (k + 1) \cdot k! )$, it suffices to show that:

2 \cdot k! \leq (k + 1) \cdot k!.

Dividing both sides by $( k! )$ (which is positive for $( k \geq 4 )$), we get:

2 \leq k + 1.

This inequality holds for all $( k \geq 1 )$, so it certainly holds for $( k \geq 4 )$. Therefore,

2^{k+1} < (k + 1)!.

Conclusion

Since we have shown that:

1. The inequality $( 2^n < n! )$ holds for the base case $( n = 4 )$,
2. If $( 2^k < k! )$ holds for some $( k \geq 4 )$, then $( 2^{k+1} < (k + 1)! )$ also holds,

we conclude by the principle of mathematical induction that $( 2^n < n! )$ for all $( n \geq 4 )$.

\boxed{2^n < n! \text{ for all } n \geq 4}.