Страница 333 номер 1098, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Упростить: 1) \(\frac{(n + 3)!}{(n + 1)!}\); 2) \(\frac{(n + 2)!}{(n - 1)!}\); 3) \(\left(\frac{1}{(n + 1)!} + \frac{1}{n!}\right) \cdot n!\); 4) \(\left(\frac{1}{n!} - \frac{1}{(n + 1)!}\right) \cdot n!\); 5) \(\left(\frac{1}{n!} - \frac{1}{(n + 2)!}\right) \cdot (n + 1)!\); 6) \(\left(\frac{1}{(n + 2)!} + \frac{1}{n!}\right) \cdot (n + 1)!\).
1) \(\frac{(n + 3)!}{(n + 1)!} = (n + 2)(n + 3)\);
2) \(\frac{(n + 2)!}{(n - 1)!} = (n + 2)(n + 1)n\);
3) \(\left( \frac{1}{(n + 1)!} + \frac{1}{n!} \right) \cdot n! = \frac{n!}{(n + 1)!} + \frac{n!}{n!} = \frac{1}{n + 1} + 1 = \frac{n + 2}{n + 1}\);
4) \(\left( \frac{1}{n!} - \frac{1}{(n + 1)!} \right) \cdot n! = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}\);
5) \(\left( \frac{1}{n!} - \frac{1}{(n + 2)!} \right) \cdot (n + 1)! = n + 1 - \frac{1}{n + 2} = \frac{n^{2} + 3n + 1}{n + 2}\);
6) \(\left( \frac{1}{(n + 2)!} + \frac{1}{n!} \right) \cdot (n + 1)! = \frac{1}{n + 2} + n + 1 = \frac{n^{2} + 3n + 3}{n + 2}\).
