Question

CONSIDERAM NUMERELE a=1/1*2+1/2*3+...+1/9*10 si b=(1-1/2)*(1-1/3)*...*(1-1/9). Aratati ca numarul 10ab este numar natural.

Answer 1

$a = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + ... + \frac{1}{9 \times 10} \\ \\ a = \frac{2 - 1}{1 \times 2} + \frac{3 - 2}{2 \times 3} + ... + \frac{10 - 9}{9 \times 10} \\ \\ a = \frac{2}{1 \times 2} - \frac{1}{1 \times 2} + \frac{3}{2 \times 3} - \frac{2}{2 \times 3} + ... + \frac{10}{9 \times 10} - \frac{9}{9 \times 10} \\ \\ a = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{9} - \frac{1}{10} \\ \\ a = \frac{1}{1} - \frac{1}{10} \\ \\ a = \frac{9}{10}$

$b = (1 - \frac{1}{2} ) (1 - \frac{1}{3} ) \times ... \times ( 1 - \frac{1}{9} ) \\ \\ b = \frac{1}{2} \times \frac{2}{3} \times ... \times \frac{8}{9} \\ \\ b = \frac{1 \times 1 \times .. \times 1}{1 \times 1 \times .. \times 9} \\ \\ b = \frac{1}{9}$

$10ab = 10 \times \frac{9}{10} \times \frac{1}{9} \\ \\ = 1$