Номер 352, страница 106 - гдз по алгебре 9 класс учебник Солтан, Солтан

Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019

Авторы: Солтан Г. Н., Солтан А. Е., Жумадилова А. Ж.

Тип: Учебник

Издательство: Кокшетау

Год издания: 2019 - 2026

ISBN: 978-601-317-424-2

Рекомендовано Министерством образования и науки Республики Казахстан

III. Последовательности. 14. Метод математической индукции - номер 352, страница 106.

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            "text" => "<p><strong>352.</strong> Для последовательности $(b_n)$, заданной перечислением ее членов, докажите формулу суммы $n$ первых членов:</p><p><strong>a)</strong> $S_n = \frac{n(n+1)(2n+1)}{6}$, если $(b_n): 1^2; 2^2; 3^2; \ldots; n^2;$</p><p><strong>б)</strong> $S_n = \frac{n(4n^2-1)}{3}$, если $(b_n): 1^2; 3^2; \ldots; (2n-1)^2;$</p><p><strong>в)</strong> $S_n = \frac{2n(n+1)(2n+1)}{3}$, если $(b_n): 2^2; 4^2; \ldots; (2n)^2;$</p>"
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            "text" => "<p><strong>352.</strong> Для последовательности $(b_n)$, заданной перечислением ее членов, докажите формулу суммы $n$ первых членов:</p><p><strong>a)</strong> $S_n = \frac{n(n+1)(2n+1)}{6}$, если $(b_n): 1^2; 2^2; 3^2; \ldots; n^2;$</p><p><strong>б)</strong> $S_n = \frac{n(4n^2-1)}{3}$, если $(b_n): 1^2; 3^2; \ldots; (2n-1)^2;$</p><p><strong>в)</strong> $S_n = \frac{2n(n+1)(2n+1)}{3}$, если $(b_n): 2^2; 4^2; \ldots; (2n)^2;$</p>"
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            "text" => "<p><strong>а)</strong> Докажем формулу $S_n = \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ методом математической индукции.</p><p><em>1. База индукции (проверка для n=1):</em></p><p>При $n=1$, левая часть формулы (сумма) равна $S_1 = 1^2 = 1$.</p><p>Правая часть формулы равна $\frac{1 \cdot (1+1) \cdot (2 \cdot 1+1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1$.</p><p>Так как $1 = 1$, формула верна для $n=1$.</p><p><em>2. Индукционное предположение (предположение для n=k):</em></p><p>Предположим, что формула верна для некоторого натурального числа $n=k$, то есть:</p><p>$S_k = 1^2 + 2^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}$</p><p><em>3. Индукционный шаг (доказательство для n=k+1):</em></p><p>Докажем, что если формула верна для $n=k$, то она верна и для $n=k+1$. То есть, нам нужно доказать, что $S_{k+1} = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$.</p><p>Рассмотрим сумму $S_{k+1}$:</p><p>$S_{k+1} = 1^2 + 2^2 + \dots + k^2 + (k+1)^2 = S_k + (k+1)^2$.</p><p>Используя индукционное предположение для $S_k$, получаем:</p><p>$S_{k+1} = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$.</p><p>Приведем к общему знаменателю и вынесем общий множитель $(k+1)$:</p><p>$S_{k+1} = (k+1) \left( \frac{k(2k+1)}{6} + k+1 \right) = (k+1) \left( \frac{k(2k+1) + 6(k+1)}{6} \right)$.</p><p>$S_{k+1} = (k+1) \left( \frac{2k^2+k+6k+6}{6} \right) = \frac{(k+1)(2k^2+7k+6)}{6}$.</p><p>Разложим квадратный трехчлен $2k^2+7k+6$ на множители: $2k^2+7k+6 = (k+2)(2k+3)$.</p><p>Подставим разложение в выражение для $S_{k+1}$:</p><p>$S_{k+1} = \frac{(k+1)(k+2)(2k+3)}{6}$.</p><p>Это и есть формула для $n=k+1$. Индукционный шаг доказан.</p><p>Таким образом, по принципу математической индукции, формула верна для любого натурального $n$.</p><p><strong>Ответ:</strong> Формула доказана.</p><p><strong>б)</strong> Докажем формулу $S_n = \sum_{k=1}^{n} (2k-1)^2 = \frac{n(4n^2-1)}{3}$ методом математической индукции.</p><p><em>1. База индукции (проверка для n=1):</em></p><p>При $n=1$, левая часть формулы (сумма) равна $S_1 = (2 \cdot 1-1)^2 = 1^2 = 1$.</p><p>Правая часть формулы равна $\frac{1 \cdot (4 \cdot 1^2 - 1)}{3} = \frac{1 \cdot 3}{3} = 1$.</p><p>Так как $1 = 1$, формула верна для $n=1$.</p><p><em>2. Индукционное предположение (предположение для n=k):</em></p><p>Предположим, что формула верна для некоторого натурального числа $n=k$, то есть:</p><p>$S_k = 1^2 + 3^2 + \dots + (2k-1)^2 = \frac{k(4k^2-1)}{3}$.</p><p><em>3. Индукционный шаг (доказательство для n=k+1):</em></p><p>Докажем, что если формула верна для $n=k$, то она верна и для $n=k+1$. Нам нужно доказать, что $S_{k+1} = \frac{(k+1)(4(k+1)^2-1)}{3}$.</p><p>Рассмотрим сумму $S_{k+1}$:</p><p>$S_{k+1} = 1^2 + 3^2 + \dots + (2k-1)^2 + (2(k+1)-1)^2 = S_k + (2k+1)^2$.</p><p>Используя индукционное предположение для $S_k$, получаем:</p><p>$S_{k+1} = \frac{k(4k^2-1)}{3} + (2k+1)^2$.</p><p>Используем формулу разности квадратов $4k^2-1 = (2k-1)(2k+1)$:</p><p>$S_{k+1} = \frac{k(2k-1)(2k+1)}{3} + (2k+1)^2$.</p><p>Вынесем общий множитель $(2k+1)$:</p><p>$S_{k+1} = (2k+1) \left( \frac{k(2k-1)}{3} + (2k+1) \right) = (2k+1) \left( \frac{2k^2-k + 3(2k+1)}{3} \right)$.</p><p>$S_{k+1} = (2k+1) \frac{2k^2-k+6k+3}{3} = \frac{(2k+1)(2k^2+5k+3)}{3}$.</p><p>Разложим квадратный трехчлен $2k^2+5k+3$ на множители: $2k^2+5k+3 = (k+1)(2k+3)$.</p><p>Подставим разложение в выражение для $S_{k+1}$:</p><p>$S_{k+1} = \frac{(k+1)(2k+1)(2k+3)}{3}$.</p><p>Теперь преобразуем правую часть целевой формулы для $n=k+1$: $\frac{(k+1)(4(k+1)^2-1)}{3}$.</p><p>$4(k+1)^2-1 = (2(k+1))^2-1^2 = (2(k+1)-1)(2(k+1)+1) = (2k+1)(2k+3)$.</p><p>Значит, правая часть равна $\frac{(k+1)(2k+1)(2k+3)}{3}$, что совпадает с нашим результатом. Индукционный шаг доказан.</p><p>Таким образом, по принципу математической индукции, формула верна для любого натурального $n$.</p><p><strong>Ответ:</strong> Формула доказана.</p><p><strong>в)</strong> Докажем формулу $S_n = \sum_{k=1}^{n} (2k)^2 = \frac{2n(n+1)(2n+1)}{3}$.</p><p><em>Способ 1: Использование результата из пункта а).</em></p><p>Сумма квадратов первых $n$ четных чисел:</p><p>$S_n = 2^2 + 4^2 + \dots + (2n)^2 = \sum_{k=1}^{n} (2k)^2 = \sum_{k=1}^{n} 4k^2$.</p><p>Вынесем константу 4 за знак суммы:</p><p>$S_n = 4 \sum_{k=1}^{n} k^2 = 4(1^2+2^2+\dots+n^2)$.</p><p>Из пункта а) известно, что $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.</p><p>Подставим эту формулу:</p><p>$S_n = 4 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}$.</p><p><em>Способ 2: Метод математической индукции.</em></p><p><em>1. База индукции (проверка для n=1):</em></p><p>При $n=1$, левая часть формулы (сумма) равна $S_1 = (2 \cdot 1)^2 = 4$.</p><p>Правая часть формулы равна $\frac{2 \cdot 1 \cdot (1+1) \cdot (2 \cdot 1+1)}{3} = \frac{2 \cdot 2 \cdot 3}{3} = 4$.</p><p>Так как $4 = 4$, формула верна для $n=1$.</p><p><em>2. Индукционное предположение (предположение для n=k):</em></p><p>Предположим, что формула верна для некоторого натурального числа $n=k$, то есть:</p><p>$S_k = 2^2 + 4^2 + \dots + (2k)^2 = \frac{2k(k+1)(2k+1)}{3}$.</p><p><em>3. Индукционный шаг (доказательство для n=k+1):</em></p><p>Докажем, что если формула верна для $n=k$, то она верна и для $n=k+1$. Нам нужно доказать, что $S_{k+1} = \frac{2(k+1)(k+2)(2k+3)}{3}$.</p><p>Рассмотрим сумму $S_{k+1}$:</p><p>$S_{k+1} = 2^2 + 4^2 + \dots + (2k)^2 + (2(k+1))^2 = S_k + 4(k+1)^2$.</p><p>Используя индукционное предположение для $S_k$, получаем:</p><p>$S_{k+1} = \frac{2k(k+1)(2k+1)}{3} + 4(k+1)^2$.</p><p>Вынесем общий множитель $2(k+1)$:</p><p>$S_{k+1} = 2(k+1) \left( \frac{k(2k+1)}{3} + 2(k+1) \right) = 2(k+1) \left( \frac{2k^2+k + 6(k+1)}{3} \right)$.</p><p>$S_{k+1} = 2(k+1) \left( \frac{2k^2+k+6k+6}{3} \right) = \frac{2(k+1)(2k^2+7k+6)}{3}$.</p><p>Как мы уже находили в пункте а), $2k^2+7k+6 = (k+2)(2k+3)$.</p><p>Подставим разложение в выражение для $S_{k+1}$:</p><p>$S_{k+1} = \frac{2(k+1)(k+2)(2k+3)}{3}$.</p><p>Это и есть формула для $n=k+1$. Индукционный шаг доказан.</p><p>Таким образом, обоими способами мы доказали, что формула верна для любого натурального $n$.</p><p><strong>Ответ:</strong> Формула доказана.</p>"
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            "text" => "<p><strong>а)</strong> Докажем формулу $S_n = \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ методом математической индукции.</p><p><em>1. База индукции (проверка для n=1):</em></p><p>При $n=1$, левая часть формулы (сумма) равна $S_1 = 1^2 = 1$.</p><p>Правая часть формулы равна $\frac{1 \cdot (1+1) \cdot (2 \cdot 1+1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1$.</p><p>Так как $1 = 1$, формула верна для $n=1$.</p><p><em>2. Индукционное предположение (предположение для n=k):</em></p><p>Предположим, что формула верна для некоторого натурального числа $n=k$, то есть:</p><p>$S_k = 1^2 + 2^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}$</p><p><em>3. Индукционный шаг (доказательство для n=k+1):</em></p><p>Докажем, что если формула верна для $n=k$, то она верна и для $n=k+1$. То есть, нам нужно доказать, что $S_{k+1} = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$.</p><p>Рассмотрим сумму $S_{k+1}$:</p><p>$S_{k+1} = 1^2 + 2^2 + \dots + k^2 + (k+1)^2 = S_k + (k+1)^2$.</p><p>Используя индукционное предположение для $S_k$, получаем:</p><p>$S_{k+1} = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$.</p><p>Приведем к общему знаменателю и вынесем общий множитель $(k+1)$:</p><p>$S_{k+1} = (k+1) \left( \frac{k(2k+1)}{6} + k+1 \right) = (k+1) \left( \frac{k(2k+1) + 6(k+1)}{6} \right)$.</p><p>$S_{k+1} = (k+1) \left( \frac{2k^2+k+6k+6}{6} \right) = \frac{(k+1)(2k^2+7k+6)}{6}$.</p><p>Разложим квадратный трехчлен $2k^2+7k+6$ на множители: $2k^2+7k+6 = (k+2)(2k+3)$.</p><p>Подставим разложение в выражение для $S_{k+1}$:</p><p>$S_{k+1} = \frac{(k+1)(k+2)(2k+3)}{6}$.</p><p>Это и есть формула для $n=k+1$. Индукционный шаг доказан.</p><p>Таким образом, по принципу математической индукции, формула верна для любого натурального $n$.</p><p><strong>Ответ:</strong> Формула доказана.</p><p><strong>б)</strong> Докажем формулу $S_n = \sum_{k=1}^{n} (2k-1)^2 = \frac{n(4n^2-1)}{3}$ методом математической индукции.</p><p><em>1. База индукции (проверка для n=1):</em></p><p>При $n=1$, левая часть формулы (сумма) равна $S_1 = (2 \cdot 1-1)^2 = 1^2 = 1$.</p><p>Правая часть формулы равна $\frac{1 \cdot (4 \cdot 1^2 - 1)}{3} = \frac{1 \cdot 3}{3} = 1$.</p><p>Так как $1 = 1$, формула верна для $n=1$.</p><p><em>2. Индукционное предположение (предположение для n=k):</em></p><p>Предположим, что формула верна для некоторого натурального числа $n=k$, то есть:</p><p>$S_k = 1^2 + 3^2 + \dots + (2k-1)^2 = \frac{k(4k^2-1)}{3}$.</p><p><em>3. Индукционный шаг (доказательство для n=k+1):</em></p><p>Докажем, что если формула верна для $n=k$, то она верна и для $n=k+1$. Нам нужно доказать, что $S_{k+1} = \frac{(k+1)(4(k+1)^2-1)}{3}$.</p><p>Рассмотрим сумму $S_{k+1}$:</p><p>$S_{k+1} = 1^2 + 3^2 + \dots + (2k-1)^2 + (2(k+1)-1)^2 = S_k + (2k+1)^2$.</p><p>Используя индукционное предположение для $S_k$, получаем:</p><p>$S_{k+1} = \frac{k(4k^2-1)}{3} + (2k+1)^2$.</p><p>Используем формулу разности квадратов $4k^2-1 = (2k-1)(2k+1)$:</p><p>$S_{k+1} = \frac{k(2k-1)(2k+1)}{3} + (2k+1)^2$.</p><p>Вынесем общий множитель $(2k+1)$:</p><p>$S_{k+1} = (2k+1) \left( \frac{k(2k-1)}{3} + (2k+1) \right) = (2k+1) \left( \frac{2k^2-k + 3(2k+1)}{3} \right)$.</p><p>$S_{k+1} = (2k+1) \frac{2k^2-k+6k+3}{3} = \frac{(2k+1)(2k^2+5k+3)}{3}$.</p><p>Разложим квадратный трехчлен $2k^2+5k+3$ на множители: $2k^2+5k+3 = (k+1)(2k+3)$.</p><p>Подставим разложение в выражение для $S_{k+1}$:</p><p>$S_{k+1} = \frac{(k+1)(2k+1)(2k+3)}{3}$.</p><p>Теперь преобразуем правую часть целевой формулы для $n=k+1$: $\frac{(k+1)(4(k+1)^2-1)}{3}$.</p><p>$4(k+1)^2-1 = (2(k+1))^2-1^2 = (2(k+1)-1)(2(k+1)+1) = (2k+1)(2k+3)$.</p><p>Значит, правая часть равна $\frac{(k+1)(2k+1)(2k+3)}{3}$, что совпадает с нашим результатом. Индукционный шаг доказан.</p><p>Таким образом, по принципу математической индукции, формула верна для любого натурального $n$.</p><p><strong>Ответ:</strong> Формула доказана.</p><p><strong>в)</strong> Докажем формулу $S_n = \sum_{k=1}^{n} (2k)^2 = \frac{2n(n+1)(2n+1)}{3}$.</p><p><em>Способ 1: Использование результата из пункта а).</em></p><p>Сумма квадратов первых $n$ четных чисел:</p><p>$S_n = 2^2 + 4^2 + \dots + (2n)^2 = \sum_{k=1}^{n} (2k)^2 = \sum_{k=1}^{n} 4k^2$.</p><p>Вынесем константу 4 за знак суммы:</p><p>$S_n = 4 \sum_{k=1}^{n} k^2 = 4(1^2+2^2+\dots+n^2)$.</p><p>Из пункта а) известно, что $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.</p><p>Подставим эту формулу:</p><p>$S_n = 4 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}$.</p><p><em>Способ 2: Метод математической индукции.</em></p><p><em>1. База индукции (проверка для n=1):</em></p><p>При $n=1$, левая часть формулы (сумма) равна $S_1 = (2 \cdot 1)^2 = 4$.</p><p>Правая часть формулы равна $\frac{2 \cdot 1 \cdot (1+1) \cdot (2 \cdot 1+1)}{3} = \frac{2 \cdot 2 \cdot 3}{3} = 4$.</p><p>Так как $4 = 4$, формула верна для $n=1$.</p><p><em>2. Индукционное предположение (предположение для n=k):</em></p><p>Предположим, что формула верна для некоторого натурального числа $n=k$, то есть:</p><p>$S_k = 2^2 + 4^2 + \dots + (2k)^2 = \frac{2k(k+1)(2k+1)}{3}$.</p><p><em>3. Индукционный шаг (доказательство для n=k+1):</em></p><p>Докажем, что если формула верна для $n=k$, то она верна и для $n=k+1$. Нам нужно доказать, что $S_{k+1} = \frac{2(k+1)(k+2)(2k+3)}{3}$.</p><p>Рассмотрим сумму $S_{k+1}$:</p><p>$S_{k+1} = 2^2 + 4^2 + \dots + (2k)^2 + (2(k+1))^2 = S_k + 4(k+1)^2$.</p><p>Используя индукционное предположение для $S_k$, получаем:</p><p>$S_{k+1} = \frac{2k(k+1)(2k+1)}{3} + 4(k+1)^2$.</p><p>Вынесем общий множитель $2(k+1)$:</p><p>$S_{k+1} = 2(k+1) \left( \frac{k(2k+1)}{3} + 2(k+1) \right) = 2(k+1) \left( \frac{2k^2+k + 6(k+1)}{3} \right)$.</p><p>$S_{k+1} = 2(k+1) \left( \frac{2k^2+k+6k+6}{3} \right) = \frac{2(k+1)(2k^2+7k+6)}{3}$.</p><p>Как мы уже находили в пункте а), $2k^2+7k+6 = (k+2)(2k+3)$.</p><p>Подставим разложение в выражение для $S_{k+1}$:</p><p>$S_{k+1} = \frac{2(k+1)(k+2)(2k+3)}{3}$.</p><p>Это и есть формула для $n=k+1$. Индукционный шаг доказан.</p><p>Таким образом, обоими способами мы доказали, что формула верна для любого натурального $n$.</p><p><strong>Ответ:</strong> Формула доказана.</p>"
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                    "created_at" => "2026-04-10 13:58:26"
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                    "created_at" => "2026-04-10 13:58:26"
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                  #table: "terms"
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                0 => App\Models\Term {#1137
                  #connection: "mysql"
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                    "created_at" => "2026-04-10 13:58:26"
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              #items: array:1 [
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                  #connection: "mysql"
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                    "created_at" => "2026-04-10 13:58:26"
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            "field_city" => Illuminate\Database\Eloquent\Collection {#1140
              #items: array:1 [
                0 => App\Models\Term {#1141
                  #connection: "mysql"
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                    "created_at" => "2026-04-10 13:58:26"
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            }
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              #escapeWhenCastingToString: false
            }
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              #items: []
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            }
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              #items: []
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            }
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              #escapeWhenCastingToString: false
            }
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            ]
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            }
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            }
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            }
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      ]
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    "page" => Illuminate\Database\Eloquent\Collection {#1166
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        0 => App\Models\BookPage {#1165
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          #keyType: "int"
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            "field_branch_parent" => Illuminate\Database\Eloquent\Collection {#1167
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            }
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              #items: array:1 [
                0 => App\Models\Book {#1168
                  #connection: "mysql"
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                    "field_subject" => Illuminate\Database\Eloquent\Collection {#1170 …2}
                    "field_class" => Illuminate\Database\Eloquent\Collection {#1171 …2}
                    "field_publisher" => Illuminate\Database\Eloquent\Collection {#1173 …2}
                    "field_author" => Illuminate\Database\Eloquent\Collection {#1175 …2}
                    "field_author_foreign" => Illuminate\Database\Eloquent\Collection {#1179 …2}
                    "field_book_type" => Illuminate\Database\Eloquent\Collection {#1180 …2}
                    "field_country" => Illuminate\Database\Eloquent\Collection {#1182 …2}
                    "field_city" => Illuminate\Database\Eloquent\Collection {#1184 …2}
                    "field_series" => Illuminate\Database\Eloquent\Collection {#1186 …2}
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№352 (с. 106)
Условие. №352 (с. 106)
скриншот условия
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 106, номер 352, Условие

352. Для последовательности $(b_n)$, заданной перечислением ее членов, докажите формулу суммы $n$ первых членов:

a) $S_n = \frac{n(n+1)(2n+1)}{6}$, если $(b_n): 1^2; 2^2; 3^2; \ldots; n^2;$

б) $S_n = \frac{n(4n^2-1)}{3}$, если $(b_n): 1^2; 3^2; \ldots; (2n-1)^2;$

в) $S_n = \frac{2n(n+1)(2n+1)}{3}$, если $(b_n): 2^2; 4^2; \ldots; (2n)^2;$

Решение. №352 (с. 106)
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 106, номер 352, Решение
Алгебра, 9 класс Учебник, авторы: Солтан Генадий Николаевич, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 106, номер 352, Решение (продолжение 2)
Решение 2 (rus). №352 (с. 106)

а) Докажем формулу $S_n = \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ методом математической индукции.

1. База индукции (проверка для n=1):

При $n=1$, левая часть формулы (сумма) равна $S_1 = 1^2 = 1$.

Правая часть формулы равна $\frac{1 \cdot (1+1) \cdot (2 \cdot 1+1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1$.

Так как $1 = 1$, формула верна для $n=1$.

2. Индукционное предположение (предположение для n=k):

Предположим, что формула верна для некоторого натурального числа $n=k$, то есть:

$S_k = 1^2 + 2^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}$

3. Индукционный шаг (доказательство для n=k+1):

Докажем, что если формула верна для $n=k$, то она верна и для $n=k+1$. То есть, нам нужно доказать, что $S_{k+1} = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$.

Рассмотрим сумму $S_{k+1}$:

$S_{k+1} = 1^2 + 2^2 + \dots + k^2 + (k+1)^2 = S_k + (k+1)^2$.

Используя индукционное предположение для $S_k$, получаем:

$S_{k+1} = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$.

Приведем к общему знаменателю и вынесем общий множитель $(k+1)$:

$S_{k+1} = (k+1) \left( \frac{k(2k+1)}{6} + k+1 \right) = (k+1) \left( \frac{k(2k+1) + 6(k+1)}{6} \right)$.

$S_{k+1} = (k+1) \left( \frac{2k^2+k+6k+6}{6} \right) = \frac{(k+1)(2k^2+7k+6)}{6}$.

Разложим квадратный трехчлен $2k^2+7k+6$ на множители: $2k^2+7k+6 = (k+2)(2k+3)$.

Подставим разложение в выражение для $S_{k+1}$:

$S_{k+1} = \frac{(k+1)(k+2)(2k+3)}{6}$.

Это и есть формула для $n=k+1$. Индукционный шаг доказан.

Таким образом, по принципу математической индукции, формула верна для любого натурального $n$.

Ответ: Формула доказана.

б) Докажем формулу $S_n = \sum_{k=1}^{n} (2k-1)^2 = \frac{n(4n^2-1)}{3}$ методом математической индукции.

1. База индукции (проверка для n=1):

При $n=1$, левая часть формулы (сумма) равна $S_1 = (2 \cdot 1-1)^2 = 1^2 = 1$.

Правая часть формулы равна $\frac{1 \cdot (4 \cdot 1^2 - 1)}{3} = \frac{1 \cdot 3}{3} = 1$.

Так как $1 = 1$, формула верна для $n=1$.

2. Индукционное предположение (предположение для n=k):

Предположим, что формула верна для некоторого натурального числа $n=k$, то есть:

$S_k = 1^2 + 3^2 + \dots + (2k-1)^2 = \frac{k(4k^2-1)}{3}$.

3. Индукционный шаг (доказательство для n=k+1):

Докажем, что если формула верна для $n=k$, то она верна и для $n=k+1$. Нам нужно доказать, что $S_{k+1} = \frac{(k+1)(4(k+1)^2-1)}{3}$.

Рассмотрим сумму $S_{k+1}$:

$S_{k+1} = 1^2 + 3^2 + \dots + (2k-1)^2 + (2(k+1)-1)^2 = S_k + (2k+1)^2$.

Используя индукционное предположение для $S_k$, получаем:

$S_{k+1} = \frac{k(4k^2-1)}{3} + (2k+1)^2$.

Используем формулу разности квадратов $4k^2-1 = (2k-1)(2k+1)$:

$S_{k+1} = \frac{k(2k-1)(2k+1)}{3} + (2k+1)^2$.

Вынесем общий множитель $(2k+1)$:

$S_{k+1} = (2k+1) \left( \frac{k(2k-1)}{3} + (2k+1) \right) = (2k+1) \left( \frac{2k^2-k + 3(2k+1)}{3} \right)$.

$S_{k+1} = (2k+1) \frac{2k^2-k+6k+3}{3} = \frac{(2k+1)(2k^2+5k+3)}{3}$.

Разложим квадратный трехчлен $2k^2+5k+3$ на множители: $2k^2+5k+3 = (k+1)(2k+3)$.

Подставим разложение в выражение для $S_{k+1}$:

$S_{k+1} = \frac{(k+1)(2k+1)(2k+3)}{3}$.

Теперь преобразуем правую часть целевой формулы для $n=k+1$: $\frac{(k+1)(4(k+1)^2-1)}{3}$.

$4(k+1)^2-1 = (2(k+1))^2-1^2 = (2(k+1)-1)(2(k+1)+1) = (2k+1)(2k+3)$.

Значит, правая часть равна $\frac{(k+1)(2k+1)(2k+3)}{3}$, что совпадает с нашим результатом. Индукционный шаг доказан.

Таким образом, по принципу математической индукции, формула верна для любого натурального $n$.

Ответ: Формула доказана.

в) Докажем формулу $S_n = \sum_{k=1}^{n} (2k)^2 = \frac{2n(n+1)(2n+1)}{3}$.

Способ 1: Использование результата из пункта а).

Сумма квадратов первых $n$ четных чисел:

$S_n = 2^2 + 4^2 + \dots + (2n)^2 = \sum_{k=1}^{n} (2k)^2 = \sum_{k=1}^{n} 4k^2$.

Вынесем константу 4 за знак суммы:

$S_n = 4 \sum_{k=1}^{n} k^2 = 4(1^2+2^2+\dots+n^2)$.

Из пункта а) известно, что $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.

Подставим эту формулу:

$S_n = 4 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}$.

Способ 2: Метод математической индукции.

1. База индукции (проверка для n=1):

При $n=1$, левая часть формулы (сумма) равна $S_1 = (2 \cdot 1)^2 = 4$.

Правая часть формулы равна $\frac{2 \cdot 1 \cdot (1+1) \cdot (2 \cdot 1+1)}{3} = \frac{2 \cdot 2 \cdot 3}{3} = 4$.

Так как $4 = 4$, формула верна для $n=1$.

2. Индукционное предположение (предположение для n=k):

Предположим, что формула верна для некоторого натурального числа $n=k$, то есть:

$S_k = 2^2 + 4^2 + \dots + (2k)^2 = \frac{2k(k+1)(2k+1)}{3}$.

3. Индукционный шаг (доказательство для n=k+1):

Докажем, что если формула верна для $n=k$, то она верна и для $n=k+1$. Нам нужно доказать, что $S_{k+1} = \frac{2(k+1)(k+2)(2k+3)}{3}$.

Рассмотрим сумму $S_{k+1}$:

$S_{k+1} = 2^2 + 4^2 + \dots + (2k)^2 + (2(k+1))^2 = S_k + 4(k+1)^2$.

Используя индукционное предположение для $S_k$, получаем:

$S_{k+1} = \frac{2k(k+1)(2k+1)}{3} + 4(k+1)^2$.

Вынесем общий множитель $2(k+1)$:

$S_{k+1} = 2(k+1) \left( \frac{k(2k+1)}{3} + 2(k+1) \right) = 2(k+1) \left( \frac{2k^2+k + 6(k+1)}{3} \right)$.

$S_{k+1} = 2(k+1) \left( \frac{2k^2+k+6k+6}{3} \right) = \frac{2(k+1)(2k^2+7k+6)}{3}$.

Как мы уже находили в пункте а), $2k^2+7k+6 = (k+2)(2k+3)$.

Подставим разложение в выражение для $S_{k+1}$:

$S_{k+1} = \frac{2(k+1)(k+2)(2k+3)}{3}$.

Это и есть формула для $n=k+1$. Индукционный шаг доказан.

Таким образом, обоими способами мы доказали, что формула верна для любого натурального $n$.

Ответ: Формула доказана.

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