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I need to solve a problem and I have the following multiplication: $\frac {3*5*7*...*2015}{2*4*6*...*2014}-1$. Is there any way to solve it or I need to let it like this?

Ghost
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2 Answers2

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Multiplying numerator and denominator by $${2*4*6*...*2014}$$ gives $$\frac{2015!}{2^{2014}\cdot(1007!)^2}-1$$ Solving this isn't easy but Wolfram gives value $\approx 35.8204856288911755012-1=34.8204856288911755012 $

LM2357
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  • I think you forgot to subtract the 1. – marty cohen Mar 17 '17 at 17:40
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    Note that $\frac{2015!}{2^{2014} \cdot (1007!)^2}$ can be rewritten as $\frac{2015}{2^{2014}} \cdot \binom{2014}{1007}$, and $\binom{2n}{n} \sim \frac{2^{2n}}{\sqrt{\pi n}}$. So we can approximate $\frac{2015!}{2^{2014} \cdot (1007!)^2} - 1$ by $\frac{2015}{\sqrt{1007 \pi}} - 1 \approx 34.82$, which is easier to compute and correct to two digits after the decimal. – Misha Lavrov Mar 17 '17 at 17:45
  • @martycohen Edited..Thanks – LM2357 Mar 17 '17 at 17:51
  • @Misha: See my answer for a more accurate approximation. – marty cohen Mar 17 '17 at 19:33
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Generalizing, and with absolutely no originality,

$\begin{array}\\ \dfrac{3*5*...*(2n+1)}{2*4*...*2n} &=\dfrac{\prod_{k=1}^n (2k+1) }{\prod_{k=1}^n (2k)}\\ &=\dfrac{\prod_{k=1}^n (2k+1) }{\prod_{k=1}^n (2k)}\dfrac{\prod_{k=1}^n (2k)}{\prod_{k=1}^n (2k)}\\ &=\dfrac{\prod_{k=1}^{2n+1} k}{(\prod_{k=1}^n (2k))^2}\\ &=\dfrac{(2n+1)!}{4^n(n!)^2}\\ &\approx\dfrac{(2n+1)\sqrt{2\pi (2n)}(\frac{2n}{e})^{2n}(1+1/(24n))}{4^n(\sqrt{2\pi n}(\frac{n}{e})^n(1+1/(12n)))^2} \qquad\text{by Stirling}\\ &=\dfrac{(2n+1)\sqrt{4\pi n}}{4^n2\pi n}\dfrac{(2n)^{2n}}{e^{2n}}\dfrac{e^{2n}}{n^{2n}}\dfrac{1+1/(24n)}{1+1/(6n)+1/(144n^2)}\\ &\approx\dfrac{(2n+1)2\sqrt{\pi n}}{2\pi n}(1+\frac1{24n})(1-\frac1{6n})\\ &=\dfrac{(2n+1)}{\sqrt{\pi n}}(1-\frac1{8n})\\ \end{array} $

For $n=1007$ this is 35.82048535271531+ which differs by about 0.0000003.

marty cohen
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