29.Calculate \( \sum_{k=1}^{\infty} 8\left(\frac{1}{2}\right)^{k+2} \)

To calculate the sum \( \sum_{k=1}^{\infty} 8\left(\frac{1}{2}\right)^{k+2} \), we start by rewriting the series: \[ \sum_{k=1}^{\infty} 8\left(\frac{1}{2}\right)^{k+2} = 8 \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{k+2} \] Now, notice that we can factor out \( \left(\frac{1}{2}\right)^{2} \): \[ = 8 \left(\frac{1}{2}\right)^{2} \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{k} \] Since \( \left(\frac{1}{2}\right)^{2} = \frac{1}{4} \): \[ = 8 \cdot \frac{1}{4} \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{k} \] Now, we need to calculate the sum \( \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{k} \). This is a geometric series where the first term \( a = \frac{1}{2} \) and the common ratio \( r = \frac{1}{2} \): \[ \sum_{k=1}^{\infty} r^k = \frac{r}{1-r} \] So we can plug in our values: \[ \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{k} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \] Now substituting back: \[ = 8 \cdot \frac{1}{4} \cdot 1 = 2 \] Thus, the sum \( \sum_{k=1}^{\infty} 8\left(\frac{1}{2}\right)^{k+2} \) is: \[ \boxed{2} \]