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$\Pi_{i=1}^{k}(2i-1)=\frac{(2k)!}{k!2^k}$

clearly the products are in the set of the natural numbers.

Step one show that P(1) is true

$2(1)-1=1$

True.

Step 2 induction assumption

$\Pi_{i=1}^{k}(2i-1)=\frac{(2k)!}{k!2^k}$

This is what we assume to be true.

Step 3

$P(k+1)$

$\Pi_{i=1}^{k+1}(2i-1)=\frac{(2k+1)!}{(k+1)!2^{k+1}}$

The right hand side can be reduced to $\frac{2}{2^k2}$

$\Pi_{i=1}^{k+1}(2i-1)=2(k+1) + \Pi_{i=1}^{k}(2i-1)$

which is the last product times the multiplication of all the previous products.

$=2k+1 + \frac{(2k)!}{k!2^k}$

$=\frac{(2k+1)(2k)!}{k!2^k}$

$=\frac{2k+1(2)}{2^k}$

I am not sure how to continue this proof.

whosleon
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Fernando Martinez
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2 Answers2

2

$\prod_{k=1}^{n+1}(2k-1)=(2(n+1)-1)\prod_{k=1}^{n}(2k-1)=(2n+1)\frac{(2n)!}{n!2^{n}}=\frac{(2(n+1))!}{2(n+1)n!2^{n}}=\frac{(2(n+1))!}{(n+1)!2^{n+1}}$

user71352
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1

Wy do you need an induction for this? It's an overkill. Instead: $$ 1 \cdot 3 \ldots (2n-1) = \frac{1 \cdot 2 \ldots 2n}{2 \cdot 4 \ldots 2n} =\frac{(2n)!}{2 \cdot (2 \cdot 2) \cdot (2 \cdot 3) \ldots (2 \cdot n)} = \frac{(2n)!}{n! 2^n} $$ The last step is because you have $n$ terms in the denominator.

Alex
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