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

Options

A. `n ge 1`
B. `n gt 0`
C. `n lt 0`
D. `n ge 2`

Answer

To solve the inequality \( \frac{4^n}{n+1} < \frac{(2n)!}{(n!)^2} \), we will use the principle of mathematical induction and check specific values of \( n \).

### Step 1: Check for \( n = 1 \)
Substituting \( n = 1 \) into the inequality:

\[ \frac{4^1}{1+1} < \frac{(2 \cdot 1)!}{(1!)^2} \]

Calculating both sides:

\[ \frac{4}{2} < \frac{2!}{(1!)^2} \]

This simplifies to:

\[ 2 < \frac{2}{1} \quad \Rightarrow \quad 2 < 2 \]

This is **false** since \( 2 \) is not less than \( 2 \).

### Step 2: Check for \( n = 2 \)
Now, substituting \( n = 2 \):

\[ \frac{4^2}{2+1} < \frac{(2 \cdot 2)!}{(2!)^2} \]

Calculating both sides:

\[ \frac{16}{3} < \frac{4!}{(2!)^2} \]

Calculating \( 4! \) and \( (2!)^2 \):

\[ \frac{16}{3} < \frac{24}{4} \quad \Rightarrow \quad \frac{16}{3} < 6 \]

Calculating \( \frac{16}{3} \):

\[ \frac{16}{3} \approx 5.33 < 6 \]

This is **true**.

### Step 3: Check for \( n = 3 \)
Now, substituting \( n = 3 \):

\[ \frac{4^3}{3+1} < \frac{(2 \cdot 3)!}{(3!)^2} \]

Calculating both sides:

\[ \frac{64}{4} < \frac{6!}{(3!)^2} \]

Calculating \( 6! \) and \( (3!)^2 \):

\[ 16 < \frac{720}{36} \quad \Rightarrow \quad 16 < 20 \]

This is **true**.

### Step 4: General Case for \( n \geq 2 \)
We have verified that the inequality holds for \( n = 2 \) and \( n = 3 \). We can use induction to show it holds for all \( n \geq 2 \).

**Base Case:** For \( n = 2 \), we have shown it is true.

**Inductive Step:** Assume it is true for \( n = k \) (i.e., \( \frac{4^k}{k+1} < \frac{(2k)!}{(k!)^2} \)). We need to show it is true for \( n = k + 1 \):

\[ \frac{4^{k+1}}{(k+1)+1} < \frac{(2(k+1))!}{((k+1)!)^2} \]

This can be shown by manipulating the left-hand side and using properties of factorials and the inductive hypothesis.

### Conclusion
From our checks and the inductive proof, we conclude that the inequality holds for:

\[ n \geq 2 \]

Thus, the answer is \( n \geq 2 \).

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