Find the sum : `(1)/(2) + (1)/(6) + (1)/(12) + (1)/(20) + (1)/(30 ) + (1)/(42) + (1)/(56) + (1)/(72) + (1)/(90) + (1)/(110) + (1)/(132)`

Options

A. `(7)/(8)`
B. `(11)/(12)`
C. `(15)/(16)`
D. `(17)/(18)`

Answer

To find the sum of the series

\[ S = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90} + \frac{1}{110} + \frac{1}{132} \]

we can first observe the denominators. The denominators can be expressed as products of consecutive integers:

- \(2 = 1 \cdot 2\)
- \(6 = 2 \cdot 3\)
- \(12 = 3 \cdot 4\)
- \(20 = 4 \cdot 5\)
- \(30 = 5 \cdot 6\)
- \(42 = 6 \cdot 7\)
- \(56 = 7 \cdot 8\)
- \(72 = 8 \cdot 9\)
- \(90 = 9 \cdot 10\)
- \(110 = 10 \cdot 11\)
- \(132 = 11 \cdot 12\)

This suggests that we can express each term as:

\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]

Thus, we can rewrite the sum \(S\) as follows:

\[ S = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \left( \frac{1}{6} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{8} \right) + \left( \frac{1}{8} - \frac{1}{9} \right) + \left( \frac{1}{9} - \frac{1}{10} \right) + \left( \frac{1}{10} - \frac{1}{11} \right) + \left( \frac{1}{11} - \frac{1}{12} \right) \]

Notice that this is a telescoping series, where most terms cancel out:

\[ S = 1 - \frac{1}{12} \]

Calculating this gives:

\[ S = 1 - \frac{1}{12} = \frac{12}{12} - \frac{1}{12} = \frac{11}{12} \]

Thus, the sum of the series is

\[ \boxed{\frac{11}{12}} \]