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.
Similar Questions
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Important Questions from Algebra
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- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
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