If `(4^(n))/(n+1)lt((2n)!)/((n!)^(2))` then `P(n)` is true for

Options

A. `nge1`
B. `ngt0`
C. `nlt0`
D. `nge2`

Answer

To solve the inequality \( \frac{4^n}{n+1} < \frac{(2n)!}{(n!)^2} \) and determine for which values of \( n \) the statement \( P(n) \) is true, we will analyze the inequality step by step.

### Step 1: Understand the Inequality
We need to analyze the inequality:
\[ \frac{4^n}{n+1} < \frac{(2n)!}{(n!)^2} \]
The right-hand side of the inequality, \( \frac{(2n)!}{(n!)^2} \), is known as the binomial coefficient \( \binom{2n}{n} \), which counts the number of ways to choose \( n \) items from \( 2n \) items.

### Step 2: Test Small Values of \( n \)
Let's test small integer values of \( n \) to find when the inequality holds.

#### For \( n = 1 \):
\[ \frac{4^1}{1+1} = \frac{4}{2} = 2 \]
\[ \frac{(2 \cdot 1)!}{(1!)^2} = \frac{2!}{1! \cdot 1!} = \frac{2}{1 \cdot 1} = 2 \]
The inequality \( 2 < 2 \) is **not true**. So, \( P(1) \) is false.

#### For \( n = 0 \):
Since \( n \) must be a non-negative integer, we cannot test \( n = 0 \) as \( 0! \) is defined but does not apply here due to the denominator \( n + 1 \).

### Step 3: Check \( n < 1 \)
Since \( n = 1 \) is not valid, we can conclude that for \( n < 1 \), the inequality does not hold.

### Step 4: Check \( n = 2 \)
Now, let's check \( n = 2 \):
\[ \frac{4^2}{2+1} = \frac{16}{3} \approx 5.33 \]
\[ \frac{(2 \cdot 2)!}{(2!)^2} = \frac{4!}{(2!)^2} = \frac{24}{4} = 6 \]
The inequality \( 5.33 < 6 \) is **true**. So, \( P(2) \) is true.

### Step 5: Check \( n = 3 \)
Now, let's check \( n = 3 \):
\[ \frac{4^3}{3+1} = \frac{64}{4} = 16 \]
\[ \frac{(2 \cdot 3)!}{(3!)^2} = \frac{6!}{(3!)^2} = \frac{720}{36} = 20 \]
The inequality \( 16 < 20 \) is **true**. So, \( P(3) \) is true.

### Step 6: General Case for \( n \geq 2 \)
To generalize, we can observe that as \( n \) increases, \( \frac{(2n)!}{(n!)^2} \) grows much faster than \( \frac{4^n}{n+1} \). Thus, we can conclude that for \( n \geq 2 \), the inequality holds true.

### Conclusion
The inequality \( P(n) \) is true for \( n \geq 2 \).