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Can you tell us what your efforts have been and where you are stuck? Posting a question without any work does not give our helpers a good place to begin to offer help other than to begin the problem, which you may have already done. I am sure you have tried to work the problem, so show our helpers what your efforts have been. This shows them where you may be going wrong, and how to best help.

- Feb 13, 2012

- 1,704

First we write the series as $\displaystyle y= \sum_{n=1}^{\infty} a_{n}$, where the $a_{n}$ are the solution of the difference equation...If $\displaystyle y=\frac{3}{4}+\frac{3*5}{4*8}+\frac{3*5*7}{4*8*12}+.....\infty$. Then $y^2+2y = $

$\displaystyle a_{n+1}= a_{n}\ \frac{2n+3}{4\ (n+1)} = \frac{a_{n}}{2}\ \{1+\frac{1}{2 (n+1)}\},\ a_{0}=1$ (1)

Following the procedure described in...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/

... You find the solution of (1)...

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

Now if You 'remember' the formula...

$\displaystyle \prod_{k=1}^{n} (1+\frac{1}{2 k}) = \frac{2}{\sqrt{\pi}}\ \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)}$ (3)

... You arrive to write...

$\displaystyle y = \frac{2}{\sqrt{\pi}}\ \sum_{n=1}^{\infty} \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)}\ 2^{-n}$ (4)

... and 'remembering' also that is...

$\displaystyle \sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{k}{2})}{\Gamma(n+1)}\ x^{n} = \frac{\Gamma(\frac{k}{2})}{(\sqrt{1-x})^{k}}$ (5)

... and...

$\displaystyle \Gamma(\frac{3}{2})= \frac{\sqrt{\pi}}{2}$ (6)

... You obtain finally...

$\displaystyle y=\frac{4}{\sqrt{\pi}} \{\frac{\Gamma(\frac{3}{2})}{(\sqrt{(\frac{1}{2}})^{3}}-\Gamma (\frac{3}{2})\}= 2\ (\sqrt{8}-1)$ (7)

Of course the problem is not trivial and I'm a little surprised that it has been proposed in the Pre-Algebra forum...

Kind regards

$\chi$ $\sigma$

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- Feb 7, 2012

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First hint: $y^2+2y = (y+1)^2 - 1$. So it will be helpful to find $y+1 = 1 + \frac{3}{4}+\frac{3*5}{4*8}+\frac{3*5*7}{4*8*12}+ \ldots.$If $\displaystyle y=\frac{3}{4}+\frac{3*5}{4*8}+\frac{3*5*7}{4*8*12}+.....\infty$. Then $y^2+2y = $

Second hint: This looks like a generalised binomial series. In fact, Newton's generalised binomial theorem states that $$(1+x)^s = 1 + \frac s1x + \frac{s(s-1)}{2!}x^2 + \frac{s(s-1)(s-2)}{3!}x^3 + \ldots$$ (and it converges provided that $|x|<1$). Can you force the series for $1+y$ into that form?

Third hint:

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- Feb 14, 2012

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I wish I can give as many thanks as possible to you for this solution, Opalg!First hint: $y^2+2y = (y+1)^2 - 1$. So it will be helpful to find $y+1 = 1 + \frac{3}{4}+\frac{3*5}{4*8}+\frac{3*5*7}{4*8*12}+ \ldots.$

Second hint: This looks like a generalised binomial series. In fact, Newton's generalised binomial theorem states that $$(1+x)^s = 1 + \frac s1x + \frac{s(s-1)}{2!}x^2 + \frac{s(s-1)(s-2)}{3!}x^3 + \ldots$$ (and it converges provided that $|x|<1$). Can you force the series for $1+y$ into that form?

Third hint:

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Yes,I wish I can give as many thanks as possible to you for this solution, Opalg!

It is nice when our members take the time to comment on this, and while our helpers do what they do not for accolades, but for the desire to simply be helpful, such comments are certainly appreciated.