Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 257

\[\boxed{\mathbf{257}.}\]

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

\[+ 2n^{2} + 3n + n^{2} + 2n + 3 =\]

\[= n\left( n^{2} + 2n + 3 \right) +\]

\[+ \left( n^{2} + 2n + 3 \right) =\]

\[= (n + 1)\left( n^{2} + 2n + 3 \right).\]

\[Пусть\ (n + 1)\ не\ кратно\ 3;\]

\[тогда\ n + 1 = 3n_{1} + 1\ или\ \]

\[n + 1 = 3n_{1} + 2.\]

\[\textbf{а)}\ n + 1 = 3n_{1} + 1:\]

\[n = 3n_{1}\]

\[\left( 9n_{1}^{2} + 6n_{1} + 3 \right)\ \vdots 3;\]

\[(n + 1)\left( n^{2} + 2n + 3 \right) \vdots 3.\]

\[\textbf{б)}\ n + 1 = 3n_{1} + 2:\]

\[n = 3n_{1} + 1\]

\[\left( 3n_{1} + 1 \right)^{2} + 2 \cdot \left( 3n_{1} + 1 \right) +\]

\[+ 3 = 9n_{1}^{2} + 6n_{1} + 1 + 6n_{1} +\]

\[+ 2 + 3 =\]

\[= 9n_{1}^{2} + 12n_{1} + 6 \vdots 3.\]

\[(n + 1)\left( n^{2} + 2n + 3 \right) \vdots 3.\]

\[Что\ и\ требовалось\ доказать.\]