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

Question

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

UpStudy AI · Accepted Answer

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.
Part a: - (i) $$ \frac{1}{3} = \frac{1}{4} + \frac{1}{12} $$ is true. - (ii) $$ \frac{1}{4} = \frac{1}{5} + \frac{1}{20} $$ is true. - (iii) $$ \frac{1}{5} = \frac{1}{6} + \frac{1}{30} $$ is true. Part b: $$ \frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)} $$

a. Check that each of the following is true.

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Solución

Part a

(i) Check

(ii) Check

(iii) Check

Part b

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