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

To solve this summation, we use the method of Partial Fractions (specifically the "Method of Differences" or Telescoping Series) to simplify the general term.

The general term is:

$$T_n = \frac{528}{n(n+1)(n+2)}$$

We can decompose this by noticing that the difference between the first and last factors in the denominator is $$(n+2) - n = 2$$. Let's rewrite the numerator:

$$T_n = \frac{528}{2} \cdot \frac{2}{n(n+1)(n+2)}$$

$$T_n = 264 \left[ \frac{(n+2) - n}{n(n+1)(n+2)} \right]$$

$$T_n = 264 \left[ \frac{n+2}{n(n+1)(n+2)} - \frac{n}{n(n+1)(n+2)} \right]$$

$$T_n = 264 \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$$

Now we sum $$T_n$$ from $$n=1$$ to $$10$$:

$$S = 264 \sum_{n=1}^{10} \left( \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right)$$

Let's write out the first few and the last terms to see how they cancel:

• For $$n=1$$: $$\left( \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} \right)$$
• For $$n=2$$: $$\left( \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} \right)$$
• ...
• For $$n=10$$: $$\left( \frac{1}{10 \cdot 11} - \frac{1}{11 \cdot 12} \right)$$

Notice that all middle terms cancel out, leaving only the first and the last:

$$S = 264 \left[ \frac{1}{1 \cdot 2} - \frac{1}{11 \cdot 12} \right]$$

$$S = 264 \left[ \frac{1}{2} - \frac{1}{132} \right]$$

Find a common denominator inside the brackets:

$$S = 264 \left[ \frac{66 - 1}{132} \right]$$

$$S = 264 \left[ \frac{65}{132} \right]$$

Since $$264 = 2 \times 132$$:

$$S = 2 \times 65 = 130$$

Final Answer: 130 (Option B)