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数列求和(五)

\[ \frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} + \frac{1}{{3 \cdot 4}} + \cdots + \frac{1}{{n \cdot \left( {n + 1} \right)}} = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}} \]

\[ \frac{1}{{1 \cdot 2 \cdot 3}} + \frac{1}{{2 \cdot 3 \cdot 4}} + \frac{1}{{3 \cdot 4 \cdot 5}} + \cdots + \frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} \] \[ = \frac{1}{4} - \frac{1}{{2\left( {n + 1} \right)\left( {n + 2} \right)}} \]

\[ \frac{1}{{1 \cdot 2 \cdot 3 \cdot 4}} + \frac{1}{{2 \cdot 3 \cdot 4 \cdot 5}} + \frac{1}{{3 \cdot 4 \cdot 5 \cdot 6}} + \cdots + \frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}} \] \[ = \frac{1}{{18}} - \frac{1}{{3\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}} \]

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

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

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

\[ \sum\limits_{j = 1}^n {\frac{1}{{\left( {2j - 1} \right)\left( {2j + 1} \right)\left( {2j + 3} \right)}}} = \frac{1}{{12}} - \frac{1}{{4\left( {2n + 1} \right)\left( {2n + 3} \right)}} \]

\[ \sum\limits_{j = 1}^n {\frac{1}{{\left( {3j - 2} \right)\left( {3j + 1} \right)\left( {3j + 4} \right)}}} = \frac{1}{{24}} - \frac{1}{{6\left( {3n + 1} \right)\left( {3n + 4} \right)}} \]




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