`1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132=?`

Options

A. `1/8`
B. `1/7`
C. `1/6`
D. `1/10`

Answer

To solve the expression \( \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90} + \frac{1}{110} + \frac{1}{132} \), we can rewrite each fraction in a form that allows us to see a pattern.

### Step-by-Step Solution:

1. **Rewrite Each Fraction**:
We can express each term in the sum as a difference of two fractions:
\[ \frac{1}{20} = \frac{1}{4} - \frac{1}{5} \]
\[ \frac{1}{30} = \frac{1}{5} - \frac{1}{6} \]
\[ \frac{1}{42} = \frac{1}{6} - \frac{1}{7} \]
\[ \frac{1}{56} = \frac{1}{7} - \frac{1}{8} \]
\[ \frac{1}{72} = \frac{1}{8} - \frac{1}{9} \]
\[ \frac{1}{90} = \frac{1}{9} - \frac{1}{10} \]
\[ \frac{1}{110} = \frac{1}{10} - \frac{1}{11} \]
\[ \frac{1}{132} = \frac{1}{11} - \frac{1}{12} \]

2. **Combine the Terms**:
Now we can substitute these expressions back into the original sum:
\[ \left(\frac{1}{4} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{8}\right) + \left(\frac{1}{8} - \frac{1}{9}\right) + \left(\frac{1}{9} - \frac{1}{10}\right) + \left(\frac{1}{10} - \frac{1}{11}\right) + \left(\frac{1}{11} - \frac{1}{12}\right) \]

3. **Simplify the Expression**:
When we combine all these terms, we notice that all intermediate terms cancel out:
\[ \frac{1}{4} - \frac{1}{12} \]

4. **Calculate the Remaining Terms**:
Now we need to compute:
\[ \frac{1}{4} - \frac{1}{12} \]
To do this, we find a common denominator, which is 12:
\[ \frac{1}{4} = \frac{3}{12} \]
Therefore:
\[ \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \]

### Final Answer:
Thus, the value of the expression \( \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90} + \frac{1}{110} + \frac{1}{132} \) is \( \frac{1}{6} \).