Prove the inequality for the indicated integer values of n.
n! > 2n, n ≥ 4
I assume you mean
n! > 2^n
4! > 2^4
for n>4,
n! = 4!*5*6*...*n > 2^4*2*2*2...*2
To prove the inequality for the given values of n, we need to show that n! (n factorial) is always greater than 2n when n is greater than or equal to 4.
First, let's expand n! to express it in terms of n:
n! = n * (n-1) * (n-2) * ... * 3 * 2 * 1
We can see that n! is a product of n consecutive positive integers. Now, let's consider 2n:
2n = 2 * n
To prove that n! > 2n for n ≥ 4, we can show that each term of n! is greater than the corresponding term of 2n.
Let's compare each term of n! with the corresponding term of 2n:
Term 1: n > 2
Term 2: (n-1) > 2
Term 3: (n-2) > 2
...
Term (n-2): 3 > 2
Term (n-1): 2 > 2
Term n: 1 > 2
From the above comparisons, we can see that for each value of n ≥ 4, all the terms from the third term onwards in n! are greater than the corresponding terms in 2n.
Now, let's consider the first two terms:
n > 2
(n-1) > 2
Since n ≥ 4, it follows that (n-1) ≥ 3. Therefore, n! is always greater than 2n for n ≥ 4.
Hence, we have proved that n! > 2n for n ≥ 4.