Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1098

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\[1)\ \frac{(n + 3)!}{(n + 1)!} =\]

\[= \frac{(n + 3) \bullet (n + 2) \bullet (n + 1)!}{(n + 1)!} =\]

\[= (n + 3)(n + 2)\]

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

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

\[= n(n + 1)(n + 2)\]

\[3)\ \left( \frac{1}{(n + 1)!} + \frac{1}{n!} \right) \bullet n! =\]

\[= \frac{n + 2}{n + 1}\]

\[4)\ \left( \frac{1}{n!} - \frac{1}{(n + 1)!} \right) \bullet n! =\]

\[= \frac{n}{n + 1}\]

\[5)\ \left( \frac{1}{n!} - \frac{1}{(n + 2)!} \right) \bullet (n + 1)! =\]

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

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

\[6)\ \left( \frac{1}{(n + 2)!} + \frac{1}{n!} \right) \bullet (n + 1)! =\]

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

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