常见数学式子-五八三

4.常见数学式子08-18

（持续更新ing…）

式子

没啥可说的，直接列式子吧（证明都在最下面）：

$1. \sum_{i = 1}^{n} i^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

$2. \sum_{1 \leq i < j \leq n} (i + j) = \frac{n (n - 1) (n + 1)}{2}$

$3. \sum_{1 \leq i \leq j \leq n} (i + j) = \frac{n^{2} (n + 1)}{2}$

$4. 对于某一等比数列，有 S_{n} = a_{1} \cdot \frac{1 - q^{n}}{1 - q} (S_{n} 为前 n 项的和， q 为公比)$

$1.$

\begin{aligned} \sum_{i = 1}^{n} i^{2} & = \sum_{i = 1}^{n} \sum_{j = 1}^{i} i \\ \sum_{i = 1}^{n} i^{2} & = \sum_{1 \leq j \leq i \leq n}^{n} i \\ \sum_{i = 1}^{n} i^{2} & = \sum_{j = 1}^{n} \sum_{i = j}^{n} i \\ \sum_{i = 1}^{n} i^{2} & = \sum_{j = 1}^{n} \frac{(n + j) (n - j + 1)}{2} \\ \sum_{i = 1}^{n} i^{2} & = \frac{1}{2} \cdot \sum_{j = 1}^{n} (n^{2} - j^{2} + n + j) \\ \sum_{i = 1}^{n} i^{2} & = \frac{1}{2} \cdot \sum_{j = 1}^{n} (n (n + 1) - j^{2} + j)) \\ 2 \cdot \sum_{i = 1}^{n} i^{2} & = \sum_{j = 1}^{n} (n (n + 1)) - \sum_{j = 1}^{n} j^{2} + \sum_{j = 1}^{n} j \\ 3 \cdot \sum_{i = 1}^{n} i^{2} & = \sum_{j = 1}^{n} (n (n + 1)) + \sum_{j = 1}^{n} j \\ 3 \cdot \sum_{i = 1}^{n} i^{2} & = n^{2} (n + 1) + \frac{n (n + 1)}{2} \\ 3 \cdot \sum_{i = 1}^{n} i^{2} & = \frac{2 n \cdot n (n + 1) + n (n + 1)}{2} \\ 3 \cdot \sum_{i = 1}^{n} i^{2} & = \frac{n (n + 1) (2 n + 1)}{2} \\ \sum_{i = 1}^{n} i^{2} & = \frac{n (n + 1) (2 n + 1)}{6} \end{aligned}

$2.$

\begin{aligned} \sum_{1 \leq i < j \leq n} (i + j) & = \sum_{i = 1}^{n} \sum_{j = 1}^{i - 1} (i + j) \\ = \sum_{i = 1}^{n} (i (i - 1) + \frac{i (i - 1)}{2}) \\ = \frac{3}{2} \cdot \sum_{i = 1}^{n} i (i - 1) \\ = \frac{3}{2} \cdot (\sum_{i = 1}^{n} i^{2} - \sum_{i = 1}^{n} i) \\ = \frac{3}{2} \cdot (\frac{n (n + 1) (2 n + 1)}{6} - \frac{n (n + 1)}{2}) \\ = \frac{3}{2} \cdot \frac{n (n + 1) (2 n - 2)}{6} \\ = \frac{n (n - 1) (n + 1)}{2} \end{aligned}

有一个地方用到了第一个式子的结论。

$3.$ 同第二个式子的证明，只不过加上了 $\frac{n (n + 1)}{2}$

$4.$

\begin{aligned} S_{n} & = a_{1} + a_{2} + \dots + a_{n - 1} + a_{n} \\ q \cdot S_{n} & = q \cdot a_{1} + q \cdot a_{2} + \dots + q \cdot a_{n - 1} + q \cdot a_{n} = S_{n + 1} \\ S_{n} - q \cdot S_{n} & = (1 - q) \cdot S_{n} = S_{n} - S_{n + 1} = a_{1} - a_{n + 1} \\ a_{n + 1} & = a_{1} \cdot q^{n} \\ S_{n} & = a_{1} \cdot \frac{1 - q^{n}}{1 - q} \end{aligned}

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本文作者： Meteor_

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