B) $$ c_n = \frac{3}{2n +4} $$
- n = 1; $$ c_1 = \frac{3}{2 \times 1 + 4} = \frac{3}{6} = \frac{1}{2} $$
- n = 2; $$ c_2 = \frac{3}{2 \times 2 + 4} = \frac{3}{8} $$
- n = 3; $$ c_3 = \frac{3}{2 \times 3 + 4} = \frac{3}{10} $$
- n = 4; $$ c_4 = \frac{3}{2 \times 4 + 4} = \frac{3}{12} = \frac{1}{4} $$
- n = 5; $$ c_5 = \frac{3}{2 \times 5 + 4} = \frac{3}{14} $$
Ответ: $$\frac{1}{2}; \frac{3}{8}; \frac{3}{10}; \frac{1}{4}; \frac{3}{14} $$.