Question

А) (1 + 1/11) × ( 1 + 1/12) × (1 + 1/13) × ... × (1 + 1/19)= б) (1 + 1/21) × (1 + 1/22) × (1 + 1/23) × ... × (1 + 1/29) =

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Answer 1

ответ $\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppppppppppppppppppppppppp\dddddd \displaystyle \large \boldsymbol{:}$

Пошаговое объяснение:

$\displaystyle \large \boldsymbol{} a) \ \ \bigg(1+ \frac{1}{11} \bigg) \cdot \bigg(1+\frac{1}{12} \bigg) \cdot \ \ldots\ \cdot \bigg(1+\frac{1}{19} \bigg)= \\\\\\ \frac{12\!\!\!\!\!\diagup}{11} \cdot \frac{13\!\!\!\!\!\diagup}{12\!\!\!\!\!\diagup} \cdot \frac{14\!\!\!\!\!\diagup}{13\!\!\!\!\!\diagup} \cdot \ \ldots \cdot \ \frac{19\!\!\!\!\!\diagup}{18\!\!\!\!\!\diagup} \cdot \frac{20}{19\!\!\!\!\!\diagup} =\boxed{\frac{20}{11} }$

$\displaystyle \large \boldsymbol{} b) \ \ \bigg(1+ \frac{1}{21} \bigg) \cdot \bigg(1+\frac{1}{22} \bigg) \cdot \ \ldots\ \cdot \bigg(1+\frac{1}{29} \bigg)= \\\\\\ \frac{22\!\!\!\!\!\diagup}{21} \cdot \frac{23\!\!\!\!\!\diagup}{22\!\!\!\!\!\diagup} \cdot \frac{24\!\!\!\!\!\diagup}{23\!\!\!\!\!\diagup} \cdot \ \ldots \cdot \ \frac{29\!\!\!\!\!\diagup}{28\!\!\!\!\!\diagup} \cdot \frac{30}{29\!\!\!\!\!\diagup} =\boxed{\frac{30}{21} }$