Step 1: The recurrence relation is \( a_n = n \cdot a_{n-1} \).
So, calculate the terms step by step: \[ a_2 = 2 \cdot a_1 = 2 \cdot 3 = 6 \] \[ a_3 = 3 \cdot a_2 = 3 \cdot 6 = 18 \] \[ a_4 = 4 \cdot a_3 = 4 \cdot 18 = 72 \] \[ a_5 = 5 \cdot a_4 = 5 \cdot 72 = 360 \] \[ a_6 = 6 \cdot a_5 = 6 \cdot 360 = 2160 \]
Let $ f(x) + 2f\left( \frac{1}{x} \right) = x^2 + 5 $ and $ 2g(x) - 3g\left( \frac{1}{2} \right) = x, \, x>0. \, \text{If} \, \alpha = \int_{1}^{2} f(x) \, dx, \, \beta = \int_{1}^{2} g(x) \, dx, \text{ then the value of } 9\alpha + \beta \text{ is:}$