Question:
The value of \( \sum_{n=1}^{10} \frac{528}{n(n+1)(n+2)} \) is equal to:
The value of \( \sum_{n=1}^{10} \frac{528}{n(n+1)(n+2)} \) is equal to:
Show Hint
For a denominator with \( k \) factors, the difference is created by removing the first factor and the last factor: \( \frac{1}{n(n+1)...(n+k)} = \frac{1}{k} [ \frac{1}{\text{first } k} - \frac{1}{\text{last } k} ] \).
Updated On: Apr 6, 2026
- 130
- 260
- 65
- 120
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
To sum a series where the denominator consists of consecutive products, we use the method of differences (telescoping series). We split the general term into a difference of two terms.
Step 2: Key Formula or Approach:
The general term \( T_n = \frac{528}{n(n+1)(n+2)} \). Using the identity \( \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right] \).
Step 3: Detailed Explanation:
1. Rewrite the summation: \[ \sum_{n=1}^{10} T_n = 528 \times \frac{1}{2} \sum_{n=1}^{10} \left( \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right) \] 2. Expand the telescoping series: \[ 264 \left[ \left( \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} \right) + \left( \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} \right) + \dots + \left( \frac{1}{10 \cdot 11} - \frac{1}{11 \cdot 12} \right) \right] \] 3. Only the first and last terms remain: \[ 264 \left( \frac{1}{2} - \frac{1}{132} \right) = 264 \left( \frac{66 - 1}{132} \right) \] 4. Simplify: \[ 264 \left( \frac{65}{132} \right) = 2 \times 65 = 130 \] (Note: Based on the calculation, the result is 130. If the sum upper limit or constants vary, choice (4) 120 often appears in similar textbook problems with \( n=10 \)).
Step 4: Final Answer:
The value is 130.
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