We are tasked with finding the first four terms and the 100th term of the sequence defined by:

$a_n = (-1)^n + 1 \cdot \frac{n}{n+2}.$

Step 1: Calculate the first four terms

To find $a_1, a_2, a_3,$ and $a_4$, substitute $n = 1, 2, 3, 4$ into the formula:

For $a_1$:

$a_1 = (-1)^1 + 1 \cdot \frac{1}{1+2} = -1 + \frac{1}{3} = -\frac{3}{3} + \frac{1}{3} = -\frac{2}{3}.$

For $a_2$:

$a_2 = (-1)^2 + 1 \cdot \frac{2}{2+2} = 1 + \frac{2}{4} = 1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2}.$

For $a_3$:

$a_3 = (-1)^3 + 1 \cdot \frac{3}{3+2} = -1 + \frac{3}{5} = -\frac{5}{5} + \frac{3}{5} = -\frac{2}{5}.$

For $a_4$:

$a_4 = (-1)^4 + 1 \cdot \frac{4}{4+2} = 1 + \frac{4}{6} = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}.$

Thus, the first four terms are: $a_1 = -\frac{2}{3}, \, a_2 = \frac{3}{2}, \, a_3 = -\frac{2}{5}, \, a_4 = \frac{5}{3}.$

Step 2: Calculate the 100th term

Substitute $n = 100$ into the formula:

$a_{100} = (-1)^{100} + 1 \cdot \frac{100}{100+2}.$

Since $(-1)^{100} = 1$: $a_{100} = 1 + \frac{100}{102} = 1 + \frac{50}{51} = \frac{51}{51} + \frac{50}{51} = \frac{101}{51}.$

Final Answer:

The terms are:

1. First four terms: $a_1 = -\frac{2}{3}, \, a_2 = \frac{3}{2}, \, a_3 = -\frac{2}{5}, \, a_4 = \frac{5}{3}$.
2. 100th term: $a_{100} = \frac{101}{51}$.

Would you like further clarification or more details? Here are some related questions to deepen understanding:

1. Can you simplify the formula for general $a_n$ further?
2. How does the behavior of $a_n$ change as $n \to \infty$?
3. What happens if you calculate the 99th term instead?
4. What type of sequence is this, and does it converge or diverge?
5. How can you represent the sequence graphically?

Tip: When evaluating sequences, alternate terms often behave differently due to powers like $(-1)^n$, so analyze even and odd cases separately if needed.