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My question is:

If $$A_{n+1} = \frac{1}{1+\frac{1}{A_n}}$$ ($n\in\mathbb{N}$) and $A_1=1$, then find the value of: $$A_1A_2 + A_2A_3 + A_3A_4 + \cdots + A_{2010} A_{2011}.$$

Please I would like to get some hints to solve this question.

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    Is that supposed to be $$A_{n+1}= \frac{1}{1+\frac{1}{A_n}}$$or $$A_{n+1}=\frac{1}{1} + \frac{1}{A_n}\ ?$$2012-06-02
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    Welcome to math.SE. In order to get the best possible answers, it is helpful if you say what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, please provide an answer to Arturo's question about your intended notation - I chose what I assumed you meant, but you need to specify the correct one.2012-06-02
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    @ArturoMagidin:The first one is correct2012-06-02
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    @ArturoMagidin:I am really sorry as the way I typed my question was very confusing.But I dont know how to write them the way u have wrote.2012-06-02
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    @user1396721: You can find guides to using LaTeX [here](http://meta.stackexchange.com/questions/68388/there-should-be-universal-latex-mathjax-guide-for-sites-supporting-it/70559#70559) and [here](http://meta.math.stackexchange.com/questions/480/math-markup-diagrams-etc-pointers-please/484#484)2012-06-02

2 Answers 2

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Here is a hint: Calculate the first few values of $A_n$; you will notice a clear pattern which you can prove to be true in general with induction. Then, note that $$\frac{1}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.$$

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    Zev Chonoles:what is induction?2012-06-02
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    @user1396721: See [the Wikipedia page](http://en.wikipedia.org/wiki/Mathematical_induction#Description) - it is a fundamental technique of proving mathematical statements. This problem should provide an excellent first experience with it.2012-06-02
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    Zev Chonoles::I observed that:A1=1 , A2=1/2 , A3=1/3...2012-06-02
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    :thus An=1/n , An+1=1/n+1.....2012-06-02
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    Excellent! Can you use induction to *prove* that, in general, $A_n=\frac{1}{n}$? Note that this involves showing that $A_1=\frac{1}{1}$ (we know that already) and then showing that, **assuming** that $A_k=\frac{1}{k}$ for some $k$, **then** we also have $A_{k+1}=\frac{1}{k+1}$.2012-06-02
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    :I did not get what you said.Sorry:(2012-06-02
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    Ok fine having aaumed that we have proved An=1/n what can be done next?2012-06-02
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    Because $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ as I noted above, we have that $$A_1A_2=\frac{1}{1}\frac{1}{2}=\frac{1}{1}-\frac{1}{2}$$ $$A_2A_3=\frac{1}{2}\frac{1}{3}=\frac{1}{2}-\frac{1}{3}$$ $$\cdots$$ $$A_{2010}A_{2011}=\frac{1}{2010}\frac{1}{2011}=\frac{1}{2010}-\frac{1}{2011}$$ so that $$A_1A_2+\cdots+A_{2010}A_{2011}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\cdots+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}$$ A lot of terms cancel :)2012-06-02
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    :Thanks a lot!!! Had great fun:)2012-06-02
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    No problem, glad to help! If my answer is satisfactory, you can click the check mark beneath the arrows on the upper left of the answer to accept it.2012-06-02
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    Good follow-through, good technique. +12012-06-02
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Notice that $A_{n+1} = (1+A_n^{-1})^{-1} = A_n/(1+A_n)$, we get $A_{n+1}^{-1} = 1 + A_n^{-1}$, and the recurrence relation $A_{n+1} = (\alpha{}A_n+\beta)/(\gamma{}A_n+\delta)$ where $\gamma\ne0$ can be solved systematically:

  1. Solve the equation $x = (\alpha{}x+\beta)/(\gamma{}x+\delta)$.
  2. If the equation has two distinct roots, say, $x_1$ and $x_2$, the sequence $\big\langle(A_n-x_1)/(A_n-x_2)\big\rangle_{n>0}$ is a geometric progression(AP). Goto 4.
  3. Otherwise, the equation has two same roots, say, $x_0$. The sequence $\big\langle(A_n-x_0)^{-1}\big\rangle$ is an arithmetic progression(GP).
  4. Find a closed-form for the AP or GP, then get the solution of the recurrence.

Some degenerate cases are not discussed, but they're trivial.