(d) Write each of the following series in sigma notation (1) \( 2+4+6+8+10+12 \) (3) \( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\text{...}+\frac{1}{128} \) (5) \( 1+8+27+64+125 \) (7) \( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\text{...} \) (to \( n \) terms)

The first series, \( 2+4+6+8+10+12 \), can be represented in sigma notation as: \(\sum_{k=1}^{6} 2k\), where \(k\) takes on the values from 1 to 6, multiplying each by 2 to obtain the even numbers. For the second series, \( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\text{...}+\frac{1}{128} \), we recognize that this is a geometric series. It can be expressed as: \(\sum_{k=1}^{7} \frac{1}{2^k}\), since \(2^1\) through \(2^7\) gives us the denominators from \(2\) to \(128\). The third series, \( 1+8+27+64+125 \), representing the cubes of integers, can be written as: \(\sum_{k=1}^{5} k^3\), since \(1^3, 2^3, 3^3, 4^3,\) and \(5^3\) yield the respective terms. Finally, the series \( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\text{...} \) (to \( n \) terms) is the harmonic series and is represented as: \(\sum_{k=1}^{n} \frac{1}{k}\), where \(k\) goes from \(1\) to \(n\).