Question

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

• A. 65
• B. 130
• C. 220
• D. 440

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to evaluate a finite sum where the general term is $\frac{528}{n(n+1)(n+2)}$.
Step 2: Key Formula or Approach:
The sum can be simplified using the method of differences (telescoping series). We need to express the general term $\frac{1}{n(n+1)(n+2)}$ as a difference of two consecutive terms.
A useful identity for this type of expression is:
\[ \frac{1}{n(n+1)...(n+k)} = \frac{1}{k} \left( \frac{1}{n(n+1)...(n+k-1)} - \frac{1}{(n+1)(n+2)...(n+k)} \right) \] For this problem, $k=2$. So,
\[ \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left( \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right) \] Step 3: Detailed Explanation:
Let the given sum be S. We can factor out the constant 528:
\[ S = 528 \sum_{n=1}^{10} \frac{1}{n(n+1)(n+2)} \] Let's use the identity for the general term:
\[ \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left( \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right) \] Let $V_n = \frac{1}{n(n+1)}$. Then the expression is $\frac{1}{2}(V_n - V_{n+1})$.
Now, the sum becomes:
\[ S = 528 \sum_{n=1}^{10} \frac{1}{2} (V_n - V_{n+1}) \] \[ S = 264 \sum_{n=1}^{10} (V_n - V_{n+1}) \] This is a telescoping sum. Let's expand it to see the cancellation:
\[ \sum_{n=1}^{10} (V_n - V_{n+1}) = (V_1 - V_2) + (V_2 - V_3) + (V_3 - V_4) + \dots + (V_{10} - V_{11}) \] All the intermediate terms cancel out, leaving:
\[ \sum_{n=1}^{10} (V_n - V_{n+1}) = V_1 - V_{11} \] Now, we calculate $V_1$ and $V_{11}$:
\[ V_1 = \frac{1}{1(1+1)} = \frac{1}{2} \] \[ V_{11} = \frac{1}{11(11+1)} = \frac{1}{11 \times 12} = \frac{1}{132} \] Substitute these values back:
\[ V_1 - V_{11} = \frac{1}{2} - \frac{1}{132} \] To subtract, find a common denominator, which is 132.
\[ V_1 - V_{11} = \frac{66}{132} - \frac{1}{132} = \frac{65}{132} \] Finally, calculate the total sum S:
\[ S = 264 \times (V_1 - V_{11}) = 264 \times \frac{65}{132} \] Since $264 = 2 \times 132$:
\[ S = 2 \times 132 \times \frac{65}{132} = 2 \times 65 = 130 \] Step 4: Final Answer:
The value of the sum is 130.