To prove the statement \( P(n) = \frac{(2n)!}{2^n \cdot n!} \) for the product of the first \( n \) odd numbers \( P(n) = 1 \times 3 \times 5 \times \cdots \times (2n - 1) \), we can follow a mathematical induction approach.
### Base Case:
First, let's check \( n = 1 \).
For \( n = 1 \):
\[ P(1) = 1 \]
According to the formula:
\[ \frac{(2 \cdot 1)!}{2^1 \cdot 1!} = \frac{2!}{2^1 \cdot 1!} = \frac{2}{2 \cdot 1} = 1 \]
Both sides of the equation are equal, so the base case holds.
### Inductive Step:
Assume the statement is true for some integer \( k \geq 1 \). That is, assume:
\[ P(k) = \frac{(2k)!}{2^k \cdot k!} \]
We need to show that the statement also holds for \( k + 1 \). That is, we need to show:
\[ P(k+1) = \frac{(2(k+1))!}{2^{k+1} \cdot (k+1)!} \]
Using the definition of \( P(n) \):
\[ P(k+1) = 1 \times 3 \times 5 \times \cdots \times (2k - 1) \times (2k + 1) \]
\[ P(k+1) = P(k) \times (2k + 1) \]
Substitute the inductive hypothesis \( P(k) = \frac{(2k)!}{2^k \cdot k!} \):
\[ P(k+1) = \frac{(2k)!}{2^k \cdot k!} \times (2k + 1) \]
We need to manipulate the expression to show it matches \( \frac{(2(k+1))!}{2^{k+1} \cdot (k+1)!} \):
\[ P(k+1) = \frac{(2k)! \times (2k + 1)}{2^k \cdot k!} \]
Recognize that:
\[ (2(k+1))! = (2k+2)! = (2k+2)(2k+1)(2k)! \]
Thus:
\[ P(k+1) = \frac{(2k+2)(2k+1)(2k)!}{2^{k+1} \cdot (k+1)!} \]
We need to simplify:
\[ P(k+1) = \frac{(2k+2)!}{2^{k+1} \cdot (k+1) \cdot k!} \]
\[ P(k+1) = \frac{(2k+2)!}{2^{k+1} \cdot (k+1)!} \]
Hence, the induction step is complete and our hypothesis is true for \( k + 1 \).
### Conclusion
By the principle of mathematical induction, we conclude that \( P(n) = \frac{(2n)!}{2^n \cdot n!} \) holds for all integers \( n \geq 1 \).
For every integer n≥1, let P(n) denote the product of the first n odd numbers:
P(n)=1×3×5×⋯×(2n−1)
Prove that for all integers n≥1, P(n)=(2n)!/2^n×n!.
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