Let's find the first five terms of the sequence given by the formula:

$a_n = \left((-1)^n + 1\right) \left(\frac{n - \frac{1}{2}}{n + 1}\right)$

We will calculate each term $a_1, a_2, a_3, a_4, a_5$ by substituting $n = 1, 2, 3, 4, 5$ into the formula:

1. For $n = 1$:

$a_1 = \left((-1)^1 + 1\right) \left(\frac{1 - \frac{1}{2}}{1 + 1}\right) = \left(-1 + 1\right) \left(\frac{\frac{1}{2}}{2}\right) = 0 \times \frac{1}{4} = 0$

2. For $n = 2$:

$a_2 = \left((-1)^2 + 1\right) \left(\frac{2 - \frac{1}{2}}{2 + 1}\right) = \left(1 + 1\right) \left(\frac{\frac{3}{2}}{3}\right) = 2 \times \frac{3}{6} = 2 \times \frac{1}{2} = 1$

3. For $n = 3$:

$a_3 = \left((-1)^3 + 1\right) \left(\frac{3 - \frac{1}{2}}{3 + 1}\right) = \left(-1 + 1\right) \left(\frac{\frac{5}{2}}{4}\right) = 0 \times \frac{5}{8} = 0$

4. For $n = 4$:

$a_4 = \left((-1)^4 + 1\right) \left(\frac{4 - \frac{1}{2}}{4 + 1}\right) = \left(1 + 1\right) \left(\frac{\frac{7}{2}}{5}\right) = 2 \times \frac{7}{10} = \frac{14}{10} = 1.4$

5. For $n = 5$:

$a_5 = \left((-1)^5 + 1\right) \left(\frac{5 - \frac{1}{2}}{5 + 1}\right) = \left(-1 + 1\right) \left(\frac{\frac{9}{2}}{6}\right) = 0 \times \frac{9}{12} = 0$

First Five Terms:

The first five terms of the sequence are:

$a_1 = 0, \quad a_2 = 1, \quad a_3 = 0, \quad a_4 = 1.4, \quad a_5 = 0$

Would you like further details or have any questions?

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Related Questions:

1. What pattern do you notice in the sequence's terms as $n$ increases?
2. How would the sequence change if $(-1)^n + 1$ were replaced by $(-1)^{n+1} + 1$?
3. Can this sequence be related to any known mathematical series?
4. What is the behavior of the sequence as $n$ approaches infinity?
5. How would the sequence change if the denominator of the fraction inside the sequence formula were $n-1$ instead of $n+1$?

Tip: Notice how sequences with alternating terms (involving $(-1)^n$) often produce terms that either cancel out or follow a repeating pattern. This can be useful in predicting future terms without direct computation.