Proving $\prod _{i=1}^{n}(2i-1)$ for all natural numbers

Sonia Ayers
2022-07-14
Answered

Proving $\prod _{i=1}^{n}(2i-1)$ for all natural numbers

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Wade Atkinson

Answered 2022-07-15
Author has **12** answers

We have,

$\prod _{{\displaystyle \phantom{\rule{thinmathspace}{0ex}}i=1}}^{{\displaystyle \phantom{\rule{thinmathspace}{0ex}}n}}(2i-1)$

$=1\cdot 3\cdot 5\cdot 7\cdots (2n-1)$

$={\displaystyle \frac{1\cdot 3\cdot 5\cdot 7\cdots (2n-1)\cdot 2\cdot 4\cdot 6\cdots 2n}{2\cdot 4\cdot 6\cdots 2n}}$

$={\displaystyle \frac{1\cdot 2\cdot 3\cdot 4\cdots \phantom{\rule{thinmathspace}{0ex}}2n}{{2}^{n}(1\cdot 2\cdot 3\cdots \phantom{\rule{thinmathspace}{0ex}}n)}}$

$={\displaystyle \frac{(2n)!}{{2}^{n}\cdot \phantom{\rule{thinmathspace}{0ex}}n!}}$

$\prod _{{\displaystyle \phantom{\rule{thinmathspace}{0ex}}i=1}}^{{\displaystyle \phantom{\rule{thinmathspace}{0ex}}n}}(2i-1)$

$=1\cdot 3\cdot 5\cdot 7\cdots (2n-1)$

$={\displaystyle \frac{1\cdot 3\cdot 5\cdot 7\cdots (2n-1)\cdot 2\cdot 4\cdot 6\cdots 2n}{2\cdot 4\cdot 6\cdots 2n}}$

$={\displaystyle \frac{1\cdot 2\cdot 3\cdot 4\cdots \phantom{\rule{thinmathspace}{0ex}}2n}{{2}^{n}(1\cdot 2\cdot 3\cdots \phantom{\rule{thinmathspace}{0ex}}n)}}$

$={\displaystyle \frac{(2n)!}{{2}^{n}\cdot \phantom{\rule{thinmathspace}{0ex}}n!}}$

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