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refactor: Improve Visuals
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‎discrete-mathematics/series-and-sequences/arithmetic-sum/arithmetic-sum.ipynb‎

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"id": "f65e306c-627b-4c98-8fe5-b6f51a985ffd",
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"metadata": {},
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"source": [
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"# Understanding Arithmetic Series Sums Through Symmetry of Averages\n",
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"# Understanding Arithmetic Sums Through Symmetry of Averages\n",
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"\n",
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"The sum of an arithmetic series is a classic result in mathematics, often proven using the method attributed to Gauss, which pairs terms from the ends of the sequence. \n",
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"\n",
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"id": "1edb5413-9527-42a7-94ad-e7fd0d59089c",
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"metadata": {},
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"source": [
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"# Average: Finding an Anchor\n",
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"# Finding the Anchor Points for Calculating the Average\n",
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"\n",
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"The **average** of any 2 points equidistant from the mid-point`a` is always `a`:\n",
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"\n",
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"$$\n",
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"\\dfrac{(a - 0) + (a + 0)}{2} = a\n",
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"\\dfrac{ \\textcolor{lightgray}{(} a \\textcolor{lightgray}{-} \\textcolor{lightgray}{0} \\textcolor{lightgray}{)} + \\textcolor{lightgray}{(} a \\textcolor{lightgray}{+} \\textcolor{lightgray}{0} \\textcolor{lightgray}{)} }{2} = a\n",
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"$$\n",
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"\n",
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"$$\n",
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"$$\n",
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"\n",
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"$$\n",
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"\\dfrac{(a - 2d) + (a + 2d)}{2} = a\n",
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"$$\n",
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"\n",
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"$$\n",
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"\\dfrac{(a - d) + (a + d)}{2} = a\n",
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"$$\n",
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"\n",
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"$$\n",
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"\\dfrac{(a - 0) + (a + 0)}{2} = a\n",
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"\\dfrac{ \\textcolor{lightgray}{(} a \\textcolor{lightgray}{-} \\textcolor{lightgray}{0} \\textcolor{lightgray}{)} + \\textcolor{lightgray}{(} a \\textcolor{lightgray}{+} \\textcolor{lightgray}{0} \\textcolor{lightgray}{)} }{2} = a\n",
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"$$\n",
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"\n",
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"Since the **first** and **last** points always exist,\n",

‎translated-notebooks/离散数学/数列与级数/等差数列求和/等差数列求和.ipynb‎

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"\n",
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"> 例如，为了求和 $1 + 2 + \\cdots + 100$，高斯将首尾项配对（如 $1 + 100$, $2 + 99$ 等）从而得出总和。\n",
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"\n",
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"在这个解释中，我们将提供一种替代方法，揭示这种配对为何有效，重点在于平均值的对称性。\n",
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"\n",
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"## 数轴上平均值的对称性\n",
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"\n",
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"让我们重新思考在一个等差数列中平均值意味着什么，在这种数列中，每一项都增加一个相同的公差。\n",
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"在这个解释中，我们将提供一种替代方法，揭示这种配对为何有效，重点在于平均值的对称性。"
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]
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},
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{
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"cell_type": "markdown",
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"id": "d5d1fe39-640f-46cb-94ea-403eed950bfc",
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"metadata": {},
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"source": [
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"# 数轴上的对称平均值\n",
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"\n",
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"- 不仅仅是“总和除以项数”，\n",
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"- 我们可以将平均值看作是一个中心点，周围的数值围绕它对称分布。\n",
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"让我们重新思考等差数列中平均值的含义，其中每一项都以相同的公差递增：\n",
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"- 不仅仅是\"总和除以项数\"\n",
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"- 我们可以将平均值视为数字对称平衡的中心点\n",
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"\n",
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"考虑任意一项：\n",
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"\n",
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"$$\n",
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"\\large\\boxed{a}\n",
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"$$\n",
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"\n",
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"如果我们向左和向右移动相同的距离 `d`，我们就可以得到两个点：\n",
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"如果我们从该项移动距离`d`，可以得到两个点：\n",
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"\n",
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"$$\n",
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"\\large{\\boxed{a - d} \\longleftarrow a \\longrightarrow \\boxed{a + d}}\n",
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"$$\n",
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"\n",
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"注意，我们构建的方式意味着 `a` 是这两个点的平均值，因为我们相对于 `a` 向左右各移动了相同距离。\n",
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"向相反方向移动相同距离时，`a`始终保持数值中心位置\n",
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"- 即**平均值**\n",
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"\n",
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"$$\n",
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"\\colorbox{lightgray}{如果我们继续以相同距离向两边移动，`a` 将始终是平均值}\n",
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"$$"
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]
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},
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{
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"cell_type": "markdown",
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"id": "c0fae9eb-c5c4-41ac-b29a-7e99e504f90f",
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"metadata": {},
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"source": [
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"# 递归构造\n",
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"\n",
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"让我们进一步说明这一点，把上面的情况想象成一个新的项 `b`：\n",
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"\n",
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"$$\n",
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"\\large{b : \\boxed{\n",
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" a - d \\longleftarrow a \\longrightarrow a + d\n",
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" }\n",
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"}\n",
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"\\large{\\textcolor{lightgray}{\\boxed{a - d}} \\textcolor{lightgray}{\\longleftarrow} a \\textcolor{lightgray}{\\longrightarrow} \\textcolor{lightgray}{\\boxed{a + d}}}\n",
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"$$\n",
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"\n",
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"从 `b` 两侧各移动 `d` 距离：\n",
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"我们可以不断延续这个模式，保持`a`作为**平均值**\n",
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"\n",
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"$$\n",
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"\\large{\n",
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" \\boxed{b - d} \\longleftarrow b \\longrightarrow \\boxed{b + d}\n",
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"}\n",
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"\\large{\\boxed{a - d} \\longleftarrow a \\longrightarrow \\boxed{a + d}}\n",
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"$$\n",
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"\n",
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"这相当于：\n",
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"\n",
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"$$\n",
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"\\large{ \\boxed{a - 2d} \\longleftarrow \\boxed{a - d} \\longleftarrow a \\longrightarrow \\boxed{a + d} \\longrightarrow \\boxed{a + 2d}}\n",
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"$$\n",
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"\n",
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"请注意，我们可以不断这样做，`a` 总是位于中心，因为我们总是以**相等的距离**但**相反的方向**移动。\n",
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"\n",
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"$$\n",
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"\\large{\n",
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" \\dots \\longleftarrow \\boxed{a - d} \\longleftarrow \\boxed{a - d} \\longleftarrow a \\longrightarrow \\boxed{a + d} \\longrightarrow \\boxed{a + 2d} \\longrightarrow \\dots\n",
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"}\n",
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"\\large{ \\textcolor{gray}{\\cdots \\boxed{a - 3d}} \\longleftarrow \\boxed{a - 2d} \\longleftarrow \\boxed{a - d} \\longleftarrow a \\longrightarrow \\boxed{a + d} \\longrightarrow \\boxed{a + 2d} \\longrightarrow \\textcolor{gray}{\\cdots \\boxed{a + 3d}}}\n",
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"$$\n",
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"\n",
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"因此，即使没有与中心等距的配对存在，只要我们选择不同的 `d`，中心始终保持不变。\n",
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"\n",
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"$$\n",
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"\\large{\n",
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" \\dots \\longleftarrow \\boxed{a - 2d} \\textcolor{darkgray}{\\longleftarrow \\boxed{a - d}} \\longleftarrow a \\longrightarrow \\textcolor{darkgray}{\\boxed{a + d} \\longrightarrow} \\boxed{a + 2d} \\longrightarrow \\dots\n",
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"}\n",
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"$$\n",
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"\n",
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"归根结底，一切都取决于你选择的 `d` 是大还是小。\n",
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"\n",
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"因此：\n",
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"因此，\n",
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"\n",
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"$$\n",
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"\\colorbox{lightgray}{中心（也是平均值）`a` 可以由任何两个相对于它对称的点推导而来}\n",
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"\\colorbox{lightgray}{如果我们持续向两个方向移动相同距离，`a`将始终保持为平均值}\n",
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"$$"
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]
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},
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"id": "c6ebcc2e-90fb-43ec-a528-6d825158dbb7",
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"source": [
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"# 平均值：寻找锚点\n",
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"# 寻找计算平均值的锚点\n",
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"\n",
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"任何两点关于中点`a`对称时，它们的**平均值**总是`a`：\n",
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"\n",
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"$$\n",
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"\\dfrac{(a - 0) + (a + 0)}{2} = a\n",
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"\\dfrac{ \\textcolor{lightgray}{(} a \\textcolor{lightgray}{-} \\textcolor{lightgray}{0} \\textcolor{lightgray}{)} + \\textcolor{lightgray}{(} a \\textcolor{lightgray}{+} \\textcolor{lightgray}{0} \\textcolor{lightgray}{)} }{2} = a\n",
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"$$\n",
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"\n",
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"$$\n",
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"\n",
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"$$\n",
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"\\dfrac{(a - 2d) + (a + 2d)}{2} = a\n",
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"$$\n",
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"\n",
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"$$\n",
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"\\dfrac{(a - d) + (a + d)}{2} = a\n",
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"$$\n",
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"\n",
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"$$\n",
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"\\dfrac{(a - 0) + (a + 0)}{2} = a\n",
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"\\dfrac{ \\textcolor{lightgray}{(} a \\textcolor{lightgray}{-} \\textcolor{lightgray}{0} \\textcolor{lightgray}{)} + \\textcolor{lightgray}{(} a \\textcolor{lightgray}{+} \\textcolor{lightgray}{0} \\textcolor{lightgray}{)} }{2} = a\n",
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"$$\n",
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"\n",
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"由于**首项**和**末项**始终存在，\n",

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