If ` f (x) = (1)/( x) - (1)/( x +1)` then what is the value of `f (1) + f( 2) + f(3) + . . . + f (10)` ?

Options

A. `(9)/(10)`
B. `(10)/(11)`
C. `(11)/(12)`
D. `(12)/(13)`

Answer

To solve the problem, we need to evaluate the function \( f(x) = \frac{1}{x} - \frac{1}{x+1} \) for values from 1 to 10 and then sum those values.

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \frac{1}{x} - \frac{1}{x+1} \]

2. **Calculate \( f(1) \)**:
\[ f(1) = \frac{1}{1} - \frac{1}{2} = 1 - 0.5 = 0.5 \]

3. **Calculate \( f(2) \)**:
\[ f(2) = \frac{1}{2} - \frac{1}{3} = 0.5 - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

4. **Calculate \( f(3) \)**:
\[ f(3) = \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12} \]

5. **Calculate \( f(4) \)**:
\[ f(4) = \frac{1}{4} - \frac{1}{5} = \frac{5}{20} - \frac{4}{20} = \frac{1}{20} \]

6. **Calculate \( f(5) \)**:
\[ f(5) = \frac{1}{5} - \frac{1}{6} = \frac{6}{30} - \frac{5}{30} = \frac{1}{30} \]

7. **Calculate \( f(6) \)**:
\[ f(6) = \frac{1}{6} - \frac{1}{7} = \frac{7}{42} - \frac{6}{42} = \frac{1}{42} \]

8. **Calculate \( f(7) \)**:
\[ f(7) = \frac{1}{7} - \frac{1}{8} = \frac{8}{56} - \frac{7}{56} = \frac{1}{56} \]

9. **Calculate \( f(8) \)**:
\[ f(8) = \frac{1}{8} - \frac{1}{9} = \frac{9}{72} - \frac{8}{72} = \frac{1}{72} \]

10. **Calculate \( f(9) \)**:
\[ f(9) = \frac{1}{9} - \frac{1}{10} = \frac{10}{90} - \frac{9}{90} = \frac{1}{90} \]

11. **Calculate \( f(10) \)**:
\[ f(10) = \frac{1}{10} - \frac{1}{11} = \frac{11}{110} - \frac{10}{110} = \frac{1}{110} \]

12. **Sum all values**:
\[ f(1) + f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) + f(9) + f(10) = 0.5 + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90} + \frac{1}{110} \]

13. **Finding a common denominator**:
The least common multiple of the denominators (1, 6, 12, 20, 30, 42, 56, 72, 90, 110) is 27720.

14. **Convert each fraction to have the common denominator and sum**:
- \( 0.5 = \frac{13860}{27720} \)
- \( \frac{1}{6} = \frac{4620}{27720} \)
- \( \frac{1}{12} = \frac{2310}{27720} \)
- \( \frac{1}{20} = \frac{1386}{27720} \)
- \( \frac{1}{30} = \frac{924}{27720} \)
- \( \frac{1}{42} = \frac{660}{27720} \)
- \( \frac{1}{56} = \frac{495}{27720} \)
- \( \frac{1}{72} = \frac{385}{27720} \)
- \( \frac{1}{90} = \frac{308}{27720} \)
- \( \frac{1}{110} = \frac{252}{27720} \)

Adding these up gives:
\[ \frac{13860 + 4620 + 2310 + 1386 + 924 + 660 + 495 + 385 + 308 + 252}{27720} = \frac{20700}{27720} = \frac{231}{308} \]

15. **Final result**:
\[ f(1) + f(2) + f(3) + \ldots + f(10) = 10 - \frac{1}{11} = \frac{110 - 1}{11} = \frac{109}{11} \approx 9.909 \]

### Final Answer:
The value of \( f(1) + f(2) + f(3) + \ldots + f(10) \) is approximately \( 9.909 \).