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The mathematical problem you've presented is to prove that the sum of a series with a pattern is equal to a specific formula:

\[ \sum_{k=1}^{n} \frac{1}{10k^2 + 20k} = \frac{n}{6n+4} \]

To solve this, let's first simplify the term inside the summation and then see if we can derive or confirm the given formula for the sum. The terms in the series are of the form:

\[ \frac{1}{10k^2 + 20k} \]

This can be factored and simplified to:

\[ \frac{1}{10k(k + 2)} \]

Using the method of partial fractions, decompose it as follows:

\[ \frac{1}{10k(k+2)} = \frac{A}{k} + \frac{B}{k+2} \]

Solving for \(A\) and \(B\),

\[ 1 = A(k+2) + Bk \]

Setting \(k = 0\), we find \(A = \frac{1}{2}\). Setting \(k = -2\), we find \(B = -\frac{1}{2}\). Thus,

\[ \frac{1}{10k(k+2)} = \frac{1}{2k} - \frac{1}{2(k+2)} \]

Now summing from \(k = 1\) to \(n\),

\[ \sum_{k=1}^{n} \left( \frac{1}{2k} - \frac{1}{2(k+2)} \right) \]

Notice that this sum telescopes:

\[ = \frac{1}{2} - \frac{1}{6} + \frac{1}{4} - \frac{1}{8} + \cdots + \frac{1}{2n} - \frac{1}{2(n+2)} \]

After cancelling out terms:

\[ = \frac{1}{2} + \frac{1}{4} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)} \]

The final terms remaining are for \(k = n+1\) and \(k = n+2\), which don't have a pair to cancel with in the series. Simplifying the remaining terms, we aim to match it with the given formula \(\frac{n}{6n+4}\).

Let's now calculate this sum and confirm whether it matches the formula given. I will do the simplification and calculation now.The simplified form of the remaining terms in the series is:

\[ \frac{n(3n + 5)}{4n^2 + 12n + 8} \]

This does not initially appear to match the given formula \(\frac{n}{6n+4}\). The discrepancy indicates that either the original summation or the partial fraction decomposition might need re-evaluation, or the problem could be more complex, requiring a different approach.

Let's review if the partial fractions or the telescoping might have been mishandled, or if there's another approach to derive the correct formula:

1. Revisit the partial fraction decomposition to ensure no algebraic errors.
2. Double-check the telescoping series calculation for any overlooked terms or misalignments.
3. Explore if a direct calculation via a different method (like integration or generating functions, if applicable) confirms or contradicts the given formula.
4. Consider possible conditions or restrictions on \(n\) that might affect the outcome.
5. Verify the simplification steps to ensure the correct mathematical manipulations were applied.

Would you like further analysis or clarification on any specific part of this process?

Prove that 1/10 + 1/40 + 1/88 + ... + 1/[(3n-1)(3n+2)] = n / (6n + 4) for all n ∈ Z.

This page provides a proof for the summation of a series that involves partial fractions and telescoping. The problem involves simplifying the series 1/[(3n-1)(3n+2)] and proving that the sum equals n / (6n + 4). Suitable for advanced high school and college-level students.

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