To simplify the given expression:

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

Let's break it down step by step.

### Step 1: Expand \(\frac{2}{n(n+1)}\)
The denominator can be expanded using partial fraction decomposition:

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

Multiply both sides by \(n(n+1)\):

\[
2 = A(n+1) + Bn
\]

Expanding the right-hand side:

\[
2 = A n + A + B n
\]

Grouping like terms:

\[
2 = (A + B)n + A
\]

To satisfy this equation, we must have:

1. \(A + B = 0\)
2. \(A = 2\)

From \(A = 2\), it follows that \(B = -2\). Thus, we can rewrite the fraction as:

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

### Step 2: Substitute this into the original expression
Now, substitute back into the original equation:

\[
\Delta T(n) = \left( \frac{2}{n} - \frac{2}{n+1} \right) - 1
\]

### Step 3: Simplify further
To combine the terms, express \(1\) as a fraction:

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

Thus, we rewrite the expression:

\[
\Delta T(n) = \left( \frac{2}{n} - \frac{2}{n+1} \right) - \frac{n(n+1)}{n(n+1)}
\]

At this point, the expression is simplified. Would you like to continue with further expansion or substitution?

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If you'd like more details or have questions, feel free to ask! Here are five related questions:

1. How does partial fraction decomposition work in other examples?
2. Can this result be used for summation series in calculus?
3. How do we apply similar techniques to higher degree polynomials?
4. What is the significance of the difference \(\Delta T(n)\)?
5. How do factorials affect simplifications in series expansions?

**Tip:** Partial fractions are a powerful tool for breaking down complex fractions into simpler, solvable parts, useful in calculus and algebra!

Learn how to simplify the expression ΔT(n) = 2/n(n+1) - 1 using partial fraction decomposition. This step-by-step solution breaks down the algebraic steps, focusing on series simplification techniques. Ideal for high school students in Grades 11-12.

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