Question

The value of the sum \[ \sum_{k=0}^{48} \frac{1}{(k + 1)(k + 2)} \] is equal to

• A. \( \frac{51}{50} \)
• B. \( \frac{51}{49} \)
• C. \( \frac{49}{50} \)
• D. \( \frac{48}{49} \)
• E. \( \frac{50}{49} \)

Hint:

For telescoping series, most terms cancel out, and you are left with only the first and last terms.

Answer 1

The Correct Option is C

We are asked to compute the sum of the series: \[ \sum_{k=0}^{48} \frac{1}{(k + 1)(k + 2)} \] This can be rewritten as: \[ \frac{1}{(k+1)(k+2)} = \frac{1}{k+1} - \frac{1}{k+2} \] Thus, the series becomes a telescoping series: \[ \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \dots + \left( \frac{1}{49} - \frac{1}{50} \right) \] All intermediate terms cancel out, and we are left with: \[ 1 - \frac{1}{50} \] Thus, the sum is: \[ 1 - \frac{1}{50} = \frac{50}{50} - \frac{1}{50} = \frac{49}{50} \] Thus, the correct answer is (C).