6) d = -2 3-4n

б) $$ d_n = \frac{-2}{3 - 4n} $$

1. n = 1; $$ d_1 = \frac{-2}{3 - 4 \times 1} = \frac{-2}{-1} = 2 $$


2. n = 2; $$ d_2 = \frac{-2}{3 - 4 \times 2} = \frac{-2}{-5} = \frac{2}{5} $$


3. n = 3; $$ d_3 = \frac{-2}{3 - 4 \times 3} = \frac{-2}{-9} = \frac{2}{9} $$


4. n = 4; $$ d_4 = \frac{-2}{3 - 4 \times 4} = \frac{-2}{-13} = \frac{2}{13} $$


5. n = 5; $$ d_5 = \frac{-2}{3 - 4 \times 5} = \frac{-2}{-17} = \frac{2}{17} $$

Ответ: $$ 2; \frac{2}{5}; \frac{2}{9}; \frac{2}{13}; \frac{2}{17} $$.