To solve the given problem, we need to evaluate the summation:
\(S = \sum_{n=1}^{10} \left( \frac{528}{n(n+1)(n+2)} \right)\)
We can simplify the term inside the summation using partial fraction decomposition.
The expression can be rewritten as:
\(\frac{528}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}\)
To find constants \(A\), \(B\), and \(C\), solve the equation:
\(528 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1)\)
Expand and compare coefficients for powers of \(n\):
- Constant term: \(2A + 2B + 1C = 0\)
- Coefficient of \(n\): \(3A + 2B + C = 0\)
- Coefficient of \(n^2\): \(A + B + C = 0\)
Solve these equations:
From \(A + B + C = 0\) and \(3A + 2B + C = 528\), we get:
\(A = 132\), \(B = 264\), \(C = -396\)
Substitute back, we have:
\(\frac{528}{n(n+1)(n+2)} = \frac{132}{n} + \frac{264}{n+1} - \frac{396}{n+2}\)
Now, the summation becomes:
\(\sum_{n=1}^{10} \left( \frac{132}{n} + \frac{264}{n+1} - \frac{396}{n+2} \right)\)
This is a telescoping series, and many terms will cancel each other. Calculate the few remaining terms:
After simplification, we focus on surviving terms at boundaries:
The terms at \(n = 1\) and \(n = 10\) are calculated, and excess internal terms range cancel each other.
The result is \(130\).
Hence, the sum \(\sum_{n=1}^{10} \left( \frac{528}{n(n+1)(n+2)} \right)\) equals 130.
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