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数列求和(四)

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

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

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

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

\[ \sum\limits_{j = 1}^n {j\left( {j + 1} \right)^2 } = \frac{1}{{12}}n\left( {n + 1} \right)\left( {n + 2} \right)\left( {3n + 5} \right) \]

\[ \sum\limits_{j = 1}^n {j\left( {j + 1} \right)^2 } \left( {j + 2} \right) = \frac{1}{{10}}n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {2n + 3} \right) \]

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

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




参阅


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