Question

The sum of value \( \sum_{n=1}^{10} \frac{528}{n(n+1)(n+2)} \) is equal to:

• A. 220
• B. 65
• C. 130
• D. 260

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves the summation of a telescopic series. We decompose the general term into a difference of two simpler terms so that intermediate terms cancel out when summed.
Step 2: Key Formula or Approach:
The general term \( T_n = \frac{528}{n(n+1)(n+2)} \) can be split using the method of differences: \[ \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left[ \frac{(n+2) - n}{n(n+1)(n+2)} \right] = \frac{1}{2} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right] \]
Step 3: Detailed Explanation:
1. Rewrite the sum: \[ S = 528 \sum_{n=1}^{10} \frac{1}{2} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right] \] \[ S = 264 \sum_{n=1}^{10} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right] \] 2. Expand the telescopic sum: \[ S = 264 \left[ \left( \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} \right) + \left( \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} \right) + \dots + \left( \frac{1}{10 \cdot 11} - \frac{1}{11 \cdot 12} \right) \right] \] 3. Cancel intermediate terms: \[ S = 264 \left[ \frac{1}{2} - \frac{1}{132} \right] \] 4. Calculate final value: \[ S = 264 \left[ \frac{66 - 1}{132} \right] = 264 \left[ \frac{65}{132} \right] \] \[ S = 2 \times 65 = 130 \]
Step 4: Final Answer:
The sum is equal to 130.