Question

N= 9•3•300 pe (1 pe 1+2+3+..+100 + 1 pe 1+2+..+101+..+ 1 pe 1+2+..+999) • 90000

Answer 1

Răspuns:

5

Explicație pas cu pas:

$\boxed {1 + 2 + 3 + ... +n = \frac{n(n + 1)}{2}}$

$\boxed {\frac{1}{n} - \frac{1}{n + 1} = \frac{1}{n \cdot (n + 1)}}$

$S = \frac{1}{1 + 2 + 3 + ... + 100} + \frac{1}{1 + 2 + 3 + ... + 101} + ... + \frac{1}{1 + 2 + 3 + ... + 999} =$

$= \frac{1}{ \frac{100 \cdot 101}{2} } + \frac{1}{ \frac{101 \cdot 102}{2} } + ... + \frac{1}{ \frac{999 \cdot 1000}{2} }$

$= \frac{2}{100 \cdot 101} + \frac{2}{101 \cdot 102} + ... + \frac{2}{999 \cdot 1000}$

$= 2 \cdot \Big(\frac{101 - 100}{100 \cdot 101} + \frac{102 - 101}{101 \cdot 102} + ... + \frac{1000 - 999}{999 \cdot 1000} \Big)$

$= 2 \cdot \Big(\frac{1}{100} - \frac{1}{101} + \frac{1}{101} - \frac{1}{102} + ... + \frac{1}{999} - \frac{1}{1000} \Big)$

$= 2 \cdot \Big(\frac{1}{100} - \frac{1}{1000} \Big) = 2 \cdot \frac{10 - 1}{1000} = 2 \cdot \frac{9}{1000}$

$N = \frac{9 \cdot 3 \cdot 300}{S \cdot 90000} = \frac{9 \cdot 3 \cdot 300}{2 \cdot \frac{9}{1000} \cdot 90000} = \\ = \frac{9 \cdot 900}{2 \cdot 9 \cdot 90} = \frac{10}{2} = \bf 5$