Evaluate (1+2/3)*(1+2/5)*(1+2/7)*(1+2/9)

Solution:

Step 1: Simplify the given expression.

Step 1.1

Express $1$ as a fraction with a denominator that matches the adjacent term. $\left( \frac{3}{3} + \frac{2}{3} \right) \cdot \left( 1 + \frac{2}{5} \right) \cdot \left( 1 + \frac{2}{7} \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 1.2

Combine the numerators over the shared denominator. $\frac{3 + 2}{3} \cdot \left( 1 + \frac{2}{5} \right) \cdot \left( 1 + \frac{2}{7} \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 1.3

Perform the addition in the numerator. $\frac{5}{3} \cdot \left( 1 + \frac{2}{5} \right) \cdot \left( 1 + \frac{2}{7} \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 1.4

Again, express $1$ as a fraction with a suitable denominator. $\frac{5}{3} \cdot \left( \frac{5}{5} + \frac{2}{5} \right) \cdot \left( 1 + \frac{2}{7} \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 1.5

Combine the numerators over the common denominator. $\frac{5}{3} \cdot \frac{5 + 2}{5} \cdot \left( 1 + \frac{2}{7} \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 1.6

Sum the numerators. $\frac{5}{3} \cdot \frac{7}{5} \cdot \left( 1 + \frac{2}{7} \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 2: Eliminate the common factor of $5$.

Step 2.1

Remove the common factor. $\frac{\cancel{5}}{3} \cdot \frac{7}{\cancel{5}} \cdot \left( 1 + \frac{2}{7} \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 2.2

Simplify the expression. $\frac{1}{3} \cdot 7 \cdot \left( 1 + \frac{2}{7} \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 3: Combine $\frac{1}{3}$ and $7$. $\frac{7}{3} \cdot \left( 1 + \frac{2}{7} \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 4: Simplify further.

Step 4.1

Express $1$ as a fraction. $\frac{7}{3} \cdot \left( \frac{7}{7} + \frac{2}{7} \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 4.2

Combine the numerators. $\frac{7}{3} \cdot \frac{7 + 2}{7} \cdot \left( 1 + \frac{2}{9} \right)$

Step 4.3

Add the numerators. $\frac{7}{3} \cdot \frac{9}{7} \cdot \left( 1 + \frac{2}{9} \right)$

Step 5: Cancel the common factor of $7$.

Step 5.1

Remove the common factor. $\frac{\cancel{7}}{3} \cdot \frac{9}{\cancel{7}} \cdot \left( 1 + \frac{2}{9} \right)$

Step 5.2

Simplify the expression. $\frac{1}{3} \cdot 9 \cdot \left( 1 + \frac{2}{9} \right)$

Step 6: Eliminate the common factor of $3$.

Step 6.1

Extract $3$ from $9$. $\frac{1}{3} \cdot \left( 3 \cdot 3 \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 6.2

Cancel the common factor. $\frac{1}{\cancel{3}} \cdot \left( \cancel{3} \cdot 3 \right) \cdot \left( 1 + \frac{2}{9} \right)$

Step 6.3

Rewrite the simplified expression. $3 \cdot \left( 1 + \frac{2}{9} \right)$

Step 7: Further simplification.

Step 7.1

Convert $1$ to a fraction. $3 \cdot \left( \frac{9}{9} + \frac{2}{9} \right)$

Step 7.2

Combine the numerators. $3 \cdot \frac{9 + 2}{9}$

Step 7.3

Sum the numerators. $3 \cdot \frac{11}{9}$

Step 8: Cancel the common factor of $3$.

Step 8.1

Factor $3$ from $9$. $3 \cdot \frac{11}{3 \cdot 3}$

Step 8.2

Remove the common factor. $\cancel{3} \cdot \frac{11}{\cancel{3} \cdot 3}$

Step 8.3

Present the simplified expression. $\frac{11}{3}$

Step 9: Represent the result in various forms.

Exact Form: $\frac{11}{3}$ Decimal Form: $3.\overline{6}$ Mixed Number Form: $3 \frac{2}{3}$

Knowledge Notes:

To solve the given problem, we used several mathematical concepts and techniques:

1. Fraction Addition: When adding fractions with the same denominator, we simply add the numerators and keep the denominator the same.

2. Simplification: We simplified the expression by combining like terms and reducing fractions when possible.

3. Cancellation: When a number appears in both the numerator and the denominator, it can be canceled out, which is essentially dividing both by that number.

4. Multiplication of Fractions: To multiply fractions, we multiply the numerators together and the denominators together.

5. Conversion to Mixed Number: A fraction greater than one can be expressed as a mixed number, which is a whole number plus a proper fraction.

These concepts are fundamental in algebra and are often used in simplifying expressions and solving equations.