12 Calculate: \( \sum_{n=0}^{\infty} 32 \cdot \frac{1}{2^{n}} \)

To calculate the sum \( \sum_{n=0}^{\infty} 32 \cdot \frac{1}{2^{n}} \), we can recognize this as a geometric series. The general form of a geometric series is: \[ S = a + ar + ar^2 + ar^3 + \ldots \] where \( a \) is the first term and \( r \) is the common ratio. In this case, the first term \( a = 32 \) (when \( n = 0 \), \( \frac{1}{2^0} = 1 \)) and the common ratio \( r = \frac{1}{2} \). The formula for the sum of an infinite geometric series (when \( |r| < 1 \)) is given by: \[ S = \frac{a}{1 - r} \] Substituting the values for \( a \) and \( r \): \[ S = \frac{32}{1 - \frac{1}{2}} = \frac{32}{\frac{1}{2}} = 32 \cdot 2 = 64 \] Therefore, the value of the sum \( \sum_{n=0}^{\infty} 32 \cdot \frac{1}{2^{n}} \) is: \[ \boxed{64} \]