Write each of the following series in sigma notation: \( \begin{array}{ll}\text { (a) } 2+4+6+8+10+12 & \text { (b) } 1+8+27+64+125 \\ \text { (c) } \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{128} & \text { (d) } 3-6+12-24+\ldots . \text { to } n \text { terms } \\ \text { (e) } \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\ldots+\frac{24}{25} & \text { (f) } 5+5+5+5+5+\ldots . \text { to } n \text { terms }\end{array} \)

For part (a), the series \( 2 + 4 + 6 + 8 + 10 + 12 \) can be expressed in sigma notation as \( \sum_{k=1}^{6} 2k \). For part (b), the series \( 1 + 8 + 27 + 64 + 125 \) is made up of cubes, and can be written in sigma notation as \( \sum_{k=1}^{5} k^3 \). For part (c), the series \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{128} \) can be represented as \( \sum_{k=1}^{7} \frac{1}{2^k} \), as there are 7 terms, starting from \( \frac{1}{2^1} \). For part (d), the alternating series \( 3 - 6 + 12 - 24 + \ldots \) can be represented in sigma notation as \( \sum_{k=1}^{n} 3 \cdot (-2)^{k-1} \). For part (e), the series \( \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \ldots + \frac{24}{25} \) can be expressed as \( \sum_{k=1}^{24} \frac{k}{k+1} \). For part (f), the series \( 5 + 5 + 5 + 5 + 5 + \ldots \) to \( n \) terms can be succinctly written in sigma notation as \( \sum_{k=1}^{n} 5 \).