А) Периметр квадратаP=4а
a=134дм
P=4a
P=4*134дм
P=536дм
б)Если периметр P=4a то сторона будет а=Р:4
Р=288м
а=?
а=Р:4
а=288м:4
а=72м
№1.
База:
n = 1:
![\frac{1}{1 * 2} = \frac{1}{1 + 1}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B1+%2A+2%7D+%3D++%5Cfrac%7B1%7D%7B1+%2B+1%7D+)
Шаг:
Допустим, что мы доказали, что наше равенство верно для n = k, то есть
![\frac{1}{1 * 2} + \frac{1}{2 * 3} + \frac{1}{3 * 4} + ... + \frac{1}{k(k + 1)} = \frac{k}{k + 1}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B1+%2A+2%7D+%2B++%5Cfrac%7B1%7D%7B2+%2A+3%7D++%2B++%5Cfrac%7B1%7D%7B3+%2A+4%7D+%2B+...+%2B++%5Cfrac%7B1%7D%7Bk%28k+%2B+1%29%7D++%3D++%5Cfrac%7Bk%7D%7Bk+%2B+1%7D+)
, теперь докажем, что это верно для n = k + 1, то есть, что
![\frac{1}{1 * 2} + \frac{1}{2 * 3} + \frac{1}{3 * 4} + ... + \frac{1}{(k + 1)(k + 2)} = \frac{k + 1}{k + 2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1+%2A+2%7D+%2B+%5Cfrac%7B1%7D%7B2+%2A+3%7D+%2B+%5Cfrac%7B1%7D%7B3+%2A+4%7D+%2B+...+%2B+%5Cfrac%7B1%7D%7B%28k+%2B+1%29%28k+%2B+2%29%7D+%3D+%5Cfrac%7Bk+%2B+1%7D%7Bk+%2B+2%7D+)
Переход:
![\frac{1}{1 * 2} + \frac{1}{2 * 3} + \frac{1}{3 * 4} + ... + \frac{1}{k(k + 1)} + \frac{1}{(k + 1)(k + 2)}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1+%2A+2%7D+%2B+%5Cfrac%7B1%7D%7B2+%2A+3%7D+%2B+%5Cfrac%7B1%7D%7B3+%2A+4%7D+%2B+...+%2B+%5Cfrac%7B1%7D%7Bk%28k+%2B+1%29%7D+%2B+%5Cfrac%7B1%7D%7B%28k+%2B+1%29%28k+%2B+2%29%7D+)
=
![(\frac{1}{1 * 2} + \frac{1}{2 * 3} + \frac{1}{3 * 4} + ... + \frac{1}{k(k + 1)}) + \frac{1}{(k + 1)(k + 2)}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B1+%2A+2%7D+%2B+%5Cfrac%7B1%7D%7B2+%2A+3%7D+%2B+%5Cfrac%7B1%7D%7B3+%2A+4%7D+%2B+...+%2B+%5Cfrac%7B1%7D%7Bk%28k+%2B+1%29%7D%29+%2B+%5Cfrac%7B1%7D%7B%28k+%2B+1%29%28k+%2B+2%29%7D)
=
![( \frac{k}{k + 1}) + \frac{1}{(k + 1)(k + 2)} = \frac{k}{k + 1} + \frac{1}{(k + 1)(k + 2)} = \frac{k(k + 2)}{(k + 1)(k + 2)} + \frac{1}{(k + 1)(k + 2)}](https://tex.z-dn.net/?f=%28+%5Cfrac%7Bk%7D%7Bk+%2B+1%7D%29+%2B+%5Cfrac%7B1%7D%7B%28k+%2B+1%29%28k+%2B+2%29%7D+%3D+%5Cfrac%7Bk%7D%7Bk+%2B+1%7D+%2B+%5Cfrac%7B1%7D%7B%28k+%2B+1%29%28k+%2B+2%29%7D+%3D++%5Cfrac%7Bk%28k+%2B+2%29%7D%7B%28k+%2B+1%29%28k+%2B+2%29%7D+%2B+%5Cfrac%7B1%7D%7B%28k+%2B+1%29%28k+%2B+2%29%7D)
=
![\frac{k(k + 2) + 1}{(k + 1)(k + 2)} = \frac{k^{2} + 2k + 1}{(k + 1)(k + 2)} = \frac{(k + 1)^{2}}{(k + 1)(k + 2)} = \frac{k + 1}{k + 2}](https://tex.z-dn.net/?f=%5Cfrac%7Bk%28k+%2B+2%29+%2B+1%7D%7B%28k+%2B+1%29%28k+%2B+2%29%7D+%3D+%5Cfrac%7Bk%5E%7B2%7D+%2B+2k+%2B+1%7D%7B%28k+%2B+1%29%28k+%2B+2%29%7D+%3D+%5Cfrac%7B%28k+%2B+1%29%5E%7B2%7D%7D%7B%28k+%2B+1%29%28k+%2B+2%29%7D+%3D+%5Cfrac%7Bk+%2B+1%7D%7Bk+%2B+2%7D)
. Что и требовалось доказать. Значит для любого числа выполняется это равенство.