What is the answer `(1)/(6) +(1)/(12) +(1)/(20) +(1)/(30) +(1)/(42) +(1)/(56)`?

Options

A. `(11)/(24)`
B. `(3)/(8)`
C. `(7)/(16)`
D. `(13)/(28)`

Answer

To solve the expression \( \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} \), we will follow these steps:

### Step 1: Find the Least Common Multiple (LCM)
First, we need to find the least common multiple (LCM) of the denominators: 6, 12, 20, 30, 42, and 56.

- The prime factorization of each number is:
- \( 6 = 2 \times 3 \)
- \( 12 = 2^2 \times 3 \)
- \( 20 = 2^2 \times 5 \)
- \( 30 = 2 \times 3 \times 5 \)
- \( 42 = 2 \times 3 \times 7 \)
- \( 56 = 2^3 \times 7 \)

The LCM will take the highest power of each prime:
- For \( 2 \): \( 2^3 \) (from 56)
- For \( 3 \): \( 3^1 \) (from 6, 12, 30, 42)
- For \( 5 \): \( 5^1 \) (from 20, 30)
- For \( 7 \): \( 7^1 \) (from 42, 56)

Thus, the LCM is:
\[ LCM = 2^3 \times 3^1 \times 5^1 \times 7^1 = 8 \times 3 \times 5 \times 7 = 840 \]

### Step 2: Rewrite Each Fraction
Now, we rewrite each fraction with the common denominator of 840:
- \( \frac{1}{6} = \frac{140}{840} \)
- \( \frac{1}{12} = \frac{70}{840} \)
- \( \frac{1}{20} = \frac{42}{840} \)
- \( \frac{1}{30} = \frac{28}{840} \)
- \( \frac{1}{42} = \frac{20}{840} \)
- \( \frac{1}{56} = \frac{15}{840} \)

### Step 3: Add the Fractions
Now, we can add these fractions:
\[ \frac{140 + 70 + 42 + 28 + 20 + 15}{840} = \frac{315}{840} \]

### Step 4: Simplify the Result
Next, we simplify \( \frac{315}{840} \):
- Both 315 and 840 can be divided by 105:
\[ \frac{315 \div 105}{840 \div 105} = \frac{3}{8} \]

### Final Answer
Thus, the final answer is:
\[ \frac{3}{8} \]

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