Question

0,999< 1 pe 1•2+ 1 pe 2•3+..+1 pe 1999•2000<0,9999

Answer 1

$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + ... + \frac{1}{1999 \times 2000} \\ \\ = \frac{2 - 1}{1 \times 2} + \frac{3 - 2}{2 \times 3} + ... + \frac{2000 - 1999}{1999 \times 2000} \\ \\ = \frac{2}{1 \times 2} - \frac{1}{1 \times 2} + \frac{3}{2 \times 3} - \frac{2}{2 \times 3} + ... + \frac{2000}{1999 \times 2000} - \frac{2000}{1999 \times 2000} \\ \\ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{1999} \\ \\ = \frac{1}{1} - \frac{1}{1999} \\ \\ = \frac{1998}{1999 } \\ \\ = 0.9995$

adevarat