Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 406

\[1)\ 1^{2} + 2^{2} + 3^{2} + \ldots + n^{2} =\]

\[= \frac{n(n + 1)(2n + 1)}{6}\]

\[n = 1:\]

\[1^{2} = 1 = \frac{1(1 + 1)(2 \bullet 1 + 1)}{6}.\]

\[n = k + 1:\]

\[1^{2} + 2^{2} + 3^{2} + \ldots + (k + 1)^{2} =\]

\[= \frac{(k + 1)(k + 2)(2k + 2 + 1)}{6};\]

\[\frac{k(k + 1)(2k + 1)}{6} + (k + 1)^{2} =\]

\[= \frac{(k + 1)(k + 2)(2k + 3)}{6};\]

\[\frac{(k + 1)\left( 2k^{2} + k \right)}{6} + \frac{(k + 1)(6k + 6)}{6} =\]

\[= \frac{(k + 1)\left( 2k^{2} + 7k + 6 \right)}{6};\]

\[\frac{(k + 1)\left( 2k^{2} + 7k + 6 \right)}{6} =\]

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

\[Что\ и\ требовалось\ доказать.\]

\[2)\ 1^{3} + 2^{3} + 3^{3} + \ldots + n^{3} =\]

\[= \frac{(n + 1)^{2}n^{2}}{4}\]

\[n = 1:\]

\[1^{3} = 1 = \frac{(1 + 1)^{2} \bullet 1^{2}}{4}.\]

\[n = k + 1:\]

\[1^{3} + 2^{3} + 3^{3} + \ldots + (k + 1)^{3} =\]

\[= \frac{(k + 1 + 1)^{2}(k + 1)^{2}}{4};\]

\[\frac{(k + 1)^{2}k^{2}}{4} + (k + 1)^{3} =\]

\[= \frac{(k + 2)^{2}(k + 1)^{2}}{4};\]

\[\frac{(k + 1)^{2}k^{2}}{4} + \frac{(k + 1)^{2}(4k + 4)}{4} =\]

\[= \frac{(k + 1)^{2}\left( k^{2} + 4k + 4 \right)}{4};\]

\[\frac{(k + 1)^{2}\left( k^{2} + 4k + 4 \right)}{4} =\]

\[= \frac{(k + 1)^{2}\left( k^{2} + 4k + 4 \right)}{4}.\]

\[Что\ и\ требовалось\ доказать.\]

\[3)\ 1^{3} + 3^{3} + 5^{3} + \ldots + (2n - 1)^{3} =\]

\[= n^{2}\left( 2n^{2} - 1 \right)\]

\[n = 1:\]

\[(2 \bullet 1 - 1)^{3} = 1 = 1^{2} \bullet \left( 2 \bullet 1^{2} - 1 \right).\]

\[n = k + 1:\]

\[1^{3} + 3^{3} + 5^{3} + \ldots + \left( 2(k + 1) - 1 \right)^{3} =\]

\[= (k + 1)^{2}\left( 2(k + 1)^{2} - 1 \right);\]

\[k^{2}\left( 2k^{2} - 1 \right) + (2k + 1)^{3} =\]

\[= \left( k^{2} + 2k + 1 \right)\left( 2k^{2} + 4k + 1 \right);\]

\[2k^{4} - k^{2} + 8k^{3} + 12k^{2} + 6k + 1 =\]

\[= 2k^{4} + 8k^{3} + 11k^{2} + 6k + 1.\]

\[Что\ и\ требовалось\ доказать.\]

\[4)\ 1^{2} - 2^{2} + 3^{2} - 4^{2} + \ldots + ( - 1)^{n - 1} \bullet n^{2} =\]

\[= ( - 1)^{n - 1} \bullet \frac{n(n + 1)}{2}\]

\[n = 1:\]

\[( - 1)^{1 - 1} \bullet 1^{2} = 1 =\]

\[= ( - 1)^{1 - 1} \bullet \frac{1(1 + 1)}{2}.\]

\[n = k + 1:\]

\[\frac{k(k + 1)}{2} - (k + 1)^{2} =\]

\[= - \frac{(k + 1)(k + 2)}{2};\]

\[\frac{(k + 1) \bullet k}{2} - \frac{(k + 1)(2k + 2)}{2} =\]

\[= \frac{(k + 1)( - k - 2)}{2};\]

\[\frac{(k + 1)( - k - 2)}{2} =\]

\[= \frac{(k + 1)( - k - 2)}{2}.\]

\[Что\ и\ требовалось\ доказать.\]