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数列求和(六)

\[ \sum\limits_{j = 1}^n {\frac{{2j - 1}}{{j\left( {j + 1} \right)\left( {j + 2} \right)}}} = \frac{3}{4} - \frac{2}{{n + 2}} + \frac{1}{{2\left( {n + 1} \right)\left( {n + 2} \right)}} \]

\[ \sum\limits_{j = 1}^n {\frac{{j + 2}}{{j\left( {j + 1} \right)\left( {j + 3} \right)}}} \] \[ = \frac{{29}}{{36}} - \frac{1}{{n + 3}} - \frac{3}{{2\left( {n + 2} \right)\left( {n + 3} \right)}} - \frac{4}{{3\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}} \]

\[ \sum\limits_{j = 1}^n {\frac{{j2^{j - 1} }}{{\left( {j + 1} \right)\left( {j + 2} \right)}}} = \frac{{2^n }}{{n + 2}} - \frac{1}{2} \]

\[ \sum\limits_{j = 1}^n {\frac{{j^2 4^j }}{{\left( {j + 1} \right)\left( {j + 2} \right)}}} = \frac{2}{3} + \frac{{\left( {n - 1} \right)4^{n + 1} }}{{3\left( {n + 2} \right)}} \]

\[ \sum\limits_{j = 1}^n {\frac{{j + 2}}{{j\left( {j + 1} \right)2^j }}} = 1 - \frac{1}{{\left( {n + 1} \right)2^n }} \]

\[ \sum\limits_{j = 1}^n {\frac{{2j + 3}}{{j\left( {j + 1} \right)3^j }}} = 1 - \frac{1}{{\left( {n + 1} \right)3^n }} \]

\[ \sum\limits_{j = 1}^n {\frac{{\left( { - 1} \right)^{j - 1} 2^j }}{{\left[ {2^j + \left( { - 1} \right)^j } \right]\left[ {2^{j + 1} + \left( { - 1} \right)^{j + 1} } \right]}}} = \frac{1}{3}\left[ {1 + \frac{{\left( { - 1} \right)^{n + 1} }}{{2^{n + 1} + \left( { - 1} \right)^{n + 1} }}} \right] \]

\[ \sum\limits_{j = 1}^n {\frac{{b\left( {b + 1} \right) \cdots \left( {b + j - 1} \right)}}{{a\left( {a + 1} \right) \cdots \left( {a + j - 1} \right)}}} = \frac{1}{{b - a + 1}}\left[ {\frac{{b\left( {b + 1} \right) \cdots \left( {b + n} \right)}}{{a\left( {a + 1} \right) \cdots \left( {a + n - 1} \right)}} - b} \right] \] 


参阅


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