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upd. math. betterexplained. part 2 - arithmetic.
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‎blog/content/post/math-better-explained-arithmetic/index.en.md‎

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---
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title: "BetterExplained - Part 2 - Arithmetic"
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description: "Mental math tricks: road speed and work hours in a year"
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description: >-
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Numbers (N, Z, Q, R), their properties and relations, operations
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<span class="op-pairs"><span class="op-pair">+ and −</span><span class="op-pair">× and /</span><span class="op-pair">\(a^{b}\) and \(\sqrt[n]{a}\)</span></span>
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date: "2026-06-01"
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slug: "math-better-explained-arithmetic"
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tags:
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{{< accordion_controls expand="Expand all" collapse="Collapse all" >}}
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{{< accordion_group_open >}}
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### 60 km/h is 1 km per minute {.toc-heading-only}
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<details class="post-accordion">
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</div>
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</details>
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{{< accordion_group_close >}}
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## Sum from 1 to n
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{{< accordion_controls expand="Expand all" collapse="Collapse all" >}}
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{{< accordion_group_open >}}
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### We're usually just given the formula and told to memorize it {.toc-heading-only}
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<details class="post-accordion">
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<summary>We're usually just given the formula and told to memorize it</summary>
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<div>
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\[
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\text{Sum from 1 to } n = \frac{n(n+1)}{2}
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\]
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\[
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\text{Sum from 1 to 100} = \frac{100(100+1)}{2} = (50)(101) = 5050
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\]
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But to build **intuition** and really understand it, you need to **derive the formula yourself**.
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Below are four ways to derive it.
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</div>
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</details>
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### 1) Split the sequence in half and add in pairs {.toc-heading-only}
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<details class="post-accordion">
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<summary>1) Split the sequence in half and add in pairs</summary>
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<div>
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```
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1 2 3 4 5
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10 9 8 7 6
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```
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Each vertical pair has the same sum: \(1 + 10 = 2 + 9 = \ldots = n + 1\). There are \(\frac{n}{2}\) pairs.
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\[
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\text{Number of pairs} \times \text{sum of each pair} = \frac{n}{2}(n+1) = \frac{n(n+1)}{2}
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\]
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</div>
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</details>
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### 2) Two rows of numbers {.toc-heading-only}
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<details class="post-accordion">
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<summary>2) Two rows of numbers</summary>
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<div>
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```
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1 2 3 4 5 6 7 8 9 10
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10 9 8 7 6 5 4 3 2 1
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```
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Add **both** rows: every column sums to \(n + 1\), and there are \(n\) columns:
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\[
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\text{Sum of both rows} = n(n+1)
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\]
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We only want **one** row, so divide by 2:
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\[
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\frac{n(n+1)}{2}
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\]
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</div>
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</details>
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### 3) Picture a triangle {.toc-heading-only}
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<details class="post-accordion">
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<summary>3) Picture a triangle</summary>
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<div>
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```
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x
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x x
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x x x
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x x x x
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x x x x x
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```
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\(n\) rows, row \(k\) has \(k\) cells — \(1 + 2 + \ldots + n\) in total. Imagine fitting two such triangles together tooth-to-tooth to form a rectangle. **Derive the formula yourself** from that picture.
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</div>
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</details>
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### 4) Via the average — the most interesting one, in my view {.toc-heading-only}
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<details class="post-accordion">
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<summary>4) Via the average — the most interesting one, in my view</summary>
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<div>
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We all know that
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\[
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\text{average} = \frac{\text{sum}}{\text{number of items}}
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\]
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which we can rewrite as
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\[
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\text{sum} = \text{average} \times \text{number of items}
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\]
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For \(1, 2, \ldots, n\) the average is easy to eyeball from the **middle** of the range: \(\frac{n+1}{2}\). There are \(n\) items, so:
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\[
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\text{sum} = \frac{n+1}{2} \cdot n = \frac{n(n+1)}{2}
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\]
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For \(1 \ldots 100\): average \(\approx 50.5\), sum \(50.5 \times 100 = 5050\).
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</div>
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</details>
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{{< accordion_group_close >}}

‎blog/content/post/math-better-explained-arithmetic/index.md‎

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---
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title: "BetterExplained - часть 2 - Арифметика"
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description: "Ментальные трюки: скорость на дороге и рабочие часы в году"
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description: >-
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Числа (N, Z, Q, R), их свойства и отношения, операции
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<span class="op-pairs"><span class="op-pair">+ и −</span><span class="op-pair">× и /</span><span class="op-pair">\(a^{b}\) и \(\sqrt[n]{a}\)</span></span>
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date: "2026-06-01"
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slug: "math-better-explained-arithmetic"
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tags:
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{{< accordion_controls expand="Развернуть все" collapse="Свернуть все" >}}
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{{< accordion_group_open >}}
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### 60 км/ч — это 1 км в минуту {.toc-heading-only}
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<details class="post-accordion">
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</div>
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</details>
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{{< accordion_group_close >}}
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## Подсчет суммы от 1 до n
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{{< accordion_controls expand="Развернуть все" collapse="Свернуть все" >}}
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{{< accordion_group_open >}}
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### Нам обычно просто дают формулу и говорят запомнить {.toc-heading-only}
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<details class="post-accordion">
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<summary>Нам обычно просто дают формулу и говорят запомнить</summary>
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<div>
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\[
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\text{Сумма от 1 до } n = \frac{n(n+1)}{2}
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\]
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\[
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\text{Сумма от 1 до 100} = \frac{100(100+1)}{2} = (50)(101) = 5050
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\]
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Но чтобы добиться **интуиции** и понять её, надо **вывести эту формулу самому**.
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Ниже — четыре техники выведения.
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</div>
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</details>
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### 1) Делим массив пополам и складываем парами {.toc-heading-only}
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<details class="post-accordion">
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<summary>1) Делим массив пополам и складываем парами</summary>
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<div>
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```
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10 9 8 7 6
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```
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Каждая вертикальная пара даёт одну и ту же сумму: \(1 + 10 = 2 + 9 = \ldots = n + 1\). Пар ровно \(\frac{n}{2}\).
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\[
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\text{Число пар} \times \text{сумма каждой пары} = \frac{n}{2}(n+1) = \frac{n(n+1)}{2}
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\]
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</div>
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</details>
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### 2) Два ряда чисел {.toc-heading-only}
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<details class="post-accordion">
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<summary>2) Два ряда чисел</summary>
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<div>
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```
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1 2 3 4 5 6 7 8 9 10
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10 9 8 7 6 5 4 3 2 1
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```
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Складываем **оба** ряда: в каждом столбце снова \(n + 1\), столбцов \(n\) штук — итого:
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\[
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\text{Сумма двух рядов} = n(n+1)
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\]
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Нужна сумма **одного** ряда, поэтому делим на 2:
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\[
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\frac{n(n+1)}{2}
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\]
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</div>
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</details>
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### 3) Представить треугольник {.toc-heading-only}
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<details class="post-accordion">
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<summary>3) Представить треугольник</summary>
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<div>
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```
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x
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x x
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x x x
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x x x x
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x x x x x
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```
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\(n\) строк, в \(k\)-й строке \(k\) «клеток» — всего \(1 + 2 + \ldots + n\) единиц. Представьте, что складываете два таких треугольника «зуб к зубу» и получаете прямоугольник. **Формулу постройте сами** из этой картинки.
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</div>
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</details>
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### 4) Через среднее — самая интересная, на мой взгляд {.toc-heading-only}
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<details class="post-accordion">
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<summary>4) Через среднее — самая интересная, на мой взгляд</summary>
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<div>
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Все знают:
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\[
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\text{среднее} = \frac{\text{сумма}}{\text{число элементов}}
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\]
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Перепишем:
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\[
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\text{сумма} = \text{среднее} \times \text{число элементов}
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\]
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Для ряда \(1, 2, \ldots, n\) среднее можно взять «на глаз» — из **центра** ряда: это \(\frac{n+1}{2}\). Элементов \(n\) штук, значит:
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\[
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\text{сумма} = \frac{n+1}{2} \cdot n = \frac{n(n+1)}{2}
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\]
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Для \(1 \ldots 100\): среднее \(\approx 50{,}5\), сумма \(50{,}5 \times 100 = 5050\).
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</div>
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</details>
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{{< accordion_group_close >}}

‎blog/layouts/partials/article/components/details.html‎

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<div class="article-details">
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{{ if .Params.categories }}
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<header class="article-category">
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{{ range (.GetTerms "categories") }}
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<a href="{{ .RelPermalink }}" {{ with .Params.style }}style="background-color: {{ .background }}; color: {{ .color }};"{{ end }}>
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{{ .LinkTitle }}
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</a>
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{{ end }}
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</header>
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{{ end }}
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<div class="article-title-wrapper">
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<h2 class="article-title">
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<a href="{{ .RelPermalink }}">
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{{- .Title -}}
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</a>
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</h2>
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{{ with .Params.description }}
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<h3 class="article-subtitle">
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{{ . | safeHTML }}
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</h3>
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{{ end }}
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</div>
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{{ $showReadingTime := .Params.readingTime | default (.Site.Params.article.readingTime) }}
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{{ $showDate := not .Date.IsZero }}
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{{ $showFooter := or $showDate $showReadingTime }}
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{{ if $showFooter }}
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<footer class="article-time">
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{{ if $showDate }}
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{{ partial "helper/icon" "date" }}
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<time class="article-time--published">
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{{- .Date.Format (or .Site.Params.dateFormat.published "Jan 02, 2006") -}}
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</time>
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{{ end }}
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{{ if $showReadingTime }}
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{{ partial "helper/icon" "clock" }}
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<time class="article-time--reading">
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{{ T "article.readingTime" .ReadingTime }}
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</time>
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</div>
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{{ end }}
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</footer>
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{{ end }}
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{{ if .IsTranslated }}
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{{ partial "helper/icon" "language" }}
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{{ range .Translations }}
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<a href="{{ .Permalink }}" class="link">{{ .Language.LanguageName }}</a>
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{{ end }}
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</footer>
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{{ end }}
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</div>

‎blog/layouts/shortcodes/accordion_controls.html‎

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document.addEventListener("click", function (e) {
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var btn = e.target.closest(".accordion-toggle-all");
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if (!btn) return;
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var article = btn.closest(".article-content");
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if (!article) return;
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var controls = btn.closest(".accordion-controls");
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if (!controls) return;
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var group = controls.nextElementSibling;
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if (!group || !group.classList.contains("post-accordion-group")) return;
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article.querySelectorAll("details.post-accordion").forEach(function (d) {
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group.querySelectorAll("details.post-accordion").forEach(function (d) {
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‎blog/layouts/shortcodes/accordion_group_close.html‎

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</div>

‎blog/layouts/shortcodes/accordion_group_open.html‎

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<div class="post-accordion-group">

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