Найти шесть первых членов последовательности, если h_n=1/3*n+6.

\[h_{n} = \frac{1}{3}n + 6;\]

\[h_{1} = \frac{1}{3} \cdot 1 + 6 = 6\frac{1}{3};\]

\[h_{2} = \frac{1}{3} \cdot 2 + 6 = 6\frac{2}{3};\]

\[h_{3} = \frac{1}{3} \cdot 3 + 6 = 7;\]

\[h_{4} = 7 + \frac{1}{3} = 7\frac{1}{3};\ \]

\[h_{5} = 7\frac{1}{3} + \frac{1}{3} = 7\frac{2}{3}\]

\[h_{6} = \frac{1}{3} \cdot 6 + 6 = 2 + 6 = 8.\]