Question

rezolvati exercitiile astea, comper clasa VII, doar ce se vede​

Answer 1

$5. \: \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + ... + \frac{1}{m {}^{2} + m } = \frac{2019}{2020} \\ \\ \frac{1}{1 {}^{2} + 1} + \frac{1}{2 {}^{2} + 2} + ... + \frac{1}{m {}^{2} + m } = \frac{2019}{2020} \\ \\ \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + ... + \frac{1}{m(m + 1)} = \frac{2019}{2020} \\ \\ \frac{2 - 1}{1\cdot2} + \frac{3 - 2}{2\cdot3} + ... + \frac{m + 1 - m}{m(m + 1)} = \frac{2019}{2020} \\ \\\small \frac{2}{1\cdot2} - \frac{1}{1\cdot2} + \frac{3}{2\cdot3} - \frac{2}{2\cdot3} + ... + \frac{m + 1}{m(m + 1)} - \frac{m}{m(m - 1)} = \frac{2019}{2020} \\ \\ \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{m} - \frac{1}{m + 1} = \frac{2019}{2020} \\ \\ \frac{1}{1} - \frac{1}{m + 1} = \frac{2019}{2020} \\ \\ \frac{m}{m + 1} = \frac{2019}{2020} \\ \\ 2020m = 2019m + 2019 \\ 2020m - 2019m = 2019 \\\bold{ m = 2019}$

SPER CA TE-AM AJUTAT SI CU ATAT ♡