Discussion
What is the value of \( \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} \)?
Answer: C
Correct answer: \( \frac{4}{5} \)
Each denominator is a product of consecutive integers: \( 2 = 1 \times 2 \), \( 6 = 2 \times 3 \), \( 12 = 3 \times 4 \), \( 20 = 4 \times 5 \). Every term can therefore be split as \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}, \] giving \[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right). \] Everything between the ends cancels - a telescoping series - leaving \[ 1 - \frac{1}{5} = \frac{4}{5}. \] The general result is that the sum of the first \( n \) such terms is \( \dfrac{n}{n+1} \), and with \( n = 4 \) that is \( 4/5 \) directly. Adding the fractions by finding a common denominator of 60 gives \( 30 + 10 + 5 + 3 = 48 \), so \( 48/60 = 4/5 \) - correct, but slower.
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