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Δημοσιεύτηκε: **Παρ Οκτ 28, 2011 6:21 pm**

Έστω $\displaystyle{(a_n)_{n\geq 1}}$ αριθμητική πρόοδος με διαφορά $\displaystyle{\omega}$ .

Βρείτε έναν κλειστό τύπο για το

$\displaystyle{\sum_{k=1}^{2011}\frac{1}{a_ka_{k+1}a_{k+2}}.}$

Δημοσιεύτηκε: **Παρ Οκτ 28, 2011 9:25 pm**

$\displaystyle{\sum\limits_{k = 1}^{2011} {\frac{1}{{{a_k}{a_{k + 1}}{a_{k + 2}}}}} = \mathop = \limits^{{a_{k + 1}} = x} = \sum\limits_{k = 1}^{2011} {\frac{1}{{\left( {x - \omega } \right)x\left( {x + \omega } \right)}}} = \sum\limits_{k = 1}^{2011} {\frac{1}{{\left( {{x^2} - {\omega ^2}} \right)x}}} = \frac{1}{{{\omega ^2}}}\sum\limits_{k = 1}^{2011} {\left( {\frac{x}{{{x^2} - {\omega ^2}}} - \frac{1}{x}} \right)} = }$

$\displaystyle{ = \frac{1}{{{\omega ^2}}}\sum\limits_{k = 1}^{2011} {\left( {\frac{1}{2}\left( {\frac{1}{{x - \omega }} + \frac{1}{{x + \omega }}} \right) - \frac{1}{x}} \right)} = \frac{1}{{2{\omega ^2}}}\sum\limits_{k = 1}^{2011} {\left( {\frac{1}{{{a_{k + 1}} - \omega }} + \frac{1}{{{a_{k + 1}} + \omega }} - \frac{2}{{{a_{k + 1}}}}} \right)} = }$

$\displaystyle{ = \frac{1}{{2{\omega ^2}}}\sum\limits_{k = 1}^{2011} {\left( {\frac{1}{{{a_k}}} + \frac{1}{{{a_{k + 2}}}} - \frac{2}{{{a_{k + 1}}}}} \right)} = \frac{1}{{2{\omega ^2}}}\left( {\sum\limits_{k = 1}^{2011} {\frac{1}{{{a_k}}}} + \sum\limits_{k = 1}^{2011} {\frac{1}{{{a_{k + 2}}}}} - 2\sum\limits_{k = 1}^{2011} {\frac{1}{{{a_{k + 1}}}}} } \right) = \frac{1}{{2{\omega ^2}}}\left( {\sum\limits_{k = 1}^{2011} {\frac{1}{{{a_k}}}} + \sum\limits_{k = 3}^{2013} {\frac{1}{{{a_k}}}} - 2\sum\limits_{k = 2}^{2012} {\frac{1}{{{a_k}}}} } \right) = }$

$\displaystyle{ = \frac{1}{{2{\omega ^2}}}\left( { - \frac{1}{{{a_{2012}}}} - \frac{1}{{{a_{2013}}}} + \sum\limits_{k = 1}^{2013} {\frac{1}{{{a_k}}}} + \sum\limits_{k = 1}^{2013} {\frac{1}{{{a_k}}}} - \frac{1}{{{a_1}}} - \frac{1}{{{a_2}}} - 2\left( { - \frac{1}{{{a_{2013}}}} - \frac{1}{{{a_1}}} + \sum\limits_{k = 1}^{2013} {\frac{1}{{{a_k}}}} } \right)} \right) = }$

$\displaystyle{ = \frac{1}{{2{\omega ^2}}}\left( { - \frac{1}{{{a_{2012}}}} + \frac{1}{{{a_{2013}}}} + \frac{1}{{{a_1}}} - \frac{1}{{{a_2}}}} \right) = \frac{1}{{2\omega }}\left( {\frac{1}{{{a_1}{a_2}}} - \frac{1}{{{a_{2012}}{a_{2013}}}}} \right) = ...}$

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