a. Check that each of the following is true. \( \begin{array}{lll}\text { (i) } \frac{1}{3}=\frac{1}{4}+\frac{1}{3 \cdot 4} & \text { (ii) } \frac{1}{4}=\frac{1}{5}+\frac{1}{4 \cdot 5} & \text { (iii) } \frac{1}{5}=\frac{1}{6}+\frac{1}{5 \cdot 6} \\ \text { b. Based on the examples in (a), write } \frac{1}{n} \text { as a sum of two unit fractions; that is, as a sum of fractions with numerator } 1\end{array} \) \( \begin{array}{l}\text { a. (i) Start by simplifying } \frac{1}{3 \cdot 4} \\ \frac{1}{4}+\frac{1}{3 \cdot 4}=\frac{1}{4}+\square \text { (Type an integer or a simplified fraction.) } \\ \text { Now, write each fraction with a common denominator and add. Choose the correct answer below. } \\ \text { A. } \frac{1}{4}+\frac{1}{12}=\frac{2}{16} \\ \text { B. } \frac{1}{4}+\frac{1}{12}=\frac{6}{12} \\ \text { C. } \frac{1}{4}+\frac{1}{12}=\frac{16}{48}\end{array} \)

Let's start by checking each of the statements in part (a) one by one. ### Part a #### (i) Check \( \frac{1}{3} = \frac{1}{4} + \frac{1}{3 \cdot 4} \) First, we need to simplify \( \frac{1}{3 \cdot 4} \): \[ 3 \cdot 4 = 12 \implies \frac{1}{3 \cdot 4} = \frac{1}{12} \] Now, we can rewrite the equation: \[ \frac{1}{3} = \frac{1}{4} + \frac{1}{12} \] Next, we need to find a common denominator to add \( \frac{1}{4} \) and \( \frac{1}{12} \). The least common multiple of 4 and 12 is 12. Rewriting \( \frac{1}{4} \) with a denominator of 12: \[ \frac{1}{4} = \frac{3}{12} \] Now we can add: \[ \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3} \] Thus, the statement \( \frac{1}{3} = \frac{1}{4} + \frac{1}{12} \) is true. #### (ii) Check \( \frac{1}{4} = \frac{1}{5} + \frac{1}{4 \cdot 5} \) First, simplify \( \frac{1}{4 \cdot 5} \): \[ 4 \cdot 5 = 20 \implies \frac{1}{4 \cdot 5} = \frac{1}{20} \] Now, we can rewrite the equation: \[ \frac{1}{4} = \frac{1}{5} + \frac{1}{20} \] Finding a common denominator for \( \frac{1}{5} \) and \( \frac{1}{20} \) (the least common multiple is 20): Rewriting \( \frac{1}{5} \): \[ \frac{1}{5} = \frac{4}{20} \] Now we can add: \[ \frac{4}{20} + \frac{1}{20} = \frac{5}{20} = \frac{1}{4} \] Thus, the statement \( \frac{1}{4} = \frac{1}{5} + \frac{1}{20} \) is true. #### (iii) Check \( \frac{1}{5} = \frac{1}{6} + \frac{1}{5 \cdot 6} \) First, simplify \( \frac{1}{5 \cdot 6} \): \[ 5 \cdot 6 = 30 \implies \frac{1}{5 \cdot 6} = \frac{1}{30} \] Now, we can rewrite the equation: \[ \frac{1}{5} = \frac{1}{6} + \frac{1}{30} \] Finding a common denominator for \( \frac{1}{6} \) and \( \frac{1}{30} \) (the least common multiple is 30): Rewriting \( \frac{1}{6} \): \[ \frac{1}{6} = \frac{5}{30} \] Now we can add: \[ \frac{5}{30} + \frac{1}{30} = \frac{6}{30} = \frac{1}{5} \] Thus, the statement \( \frac{1}{5} = \frac{1}{6} + \frac{1}{30} \) is true. ### Part b Based on the examples in (a), we can generalize the pattern. We can express \( \frac{1}{n} \) as: \[ \frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)} \] This follows the same structure as the previous examples.