Summation of Rational Expression Series

To find the value of the sum: $S = \sum_{n=1}^{10} \left( \frac{528}{n(n+1)(n+2)} \right)$

We can express the general term $T_n$ using partial fractions via the method of differences (telescoping series).

Notice that the numerator of the fractional part can be rewritten as: $\frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left[ \frac{(n+2) - n}{n(n+1)(n+2)} \right]$

Splitting the fraction gives: $\frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$

Substituting this back into the general term $T_n$: $T_n = \frac{528}{2} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right] = 264 \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$

Now, summing $T_n$ from $n = 1$ to $n = 10$: $S = 264 \sum_{n=1}^{10} \left[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} \right]$

Expanding the sum gives a telescoping series where intermediate terms cancel out: $S = 264 \left[ \left( \frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} \right) + \left( \frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} \right) + \dots + \left( \frac{1}{10 \cdot 11} - \frac{1}{11 \cdot 12} \right) \right]$

All terms except the first and the last cancel out: $S = 264 \left( \frac{1}{1 \cdot 2} - \frac{1}{11 \cdot 12} \right)$

Simplify the expression inside the parentheses: $S = 264 \left( \frac{1}{2} - \frac{1}{132} \right)$ $S = 264 \left( \frac{66 - 1}{132} \right) = 264 \left( \frac{65}{132} \right)$

Since $\frac{264}{132} = 2$: $S = 2 \times 65 = 130$

Hence, the correct option is B ($130$).