Find the first five terms of the sequence a sub n=(-1)^n+1 (n-1)/2n+1.

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

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

We'll 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 \( n = 1 \):**

\[
a_1 = \frac{(-1)^{1+1} \cdot (1-1)}{2(1) + 1} = \frac{(-1)^2 \cdot 0}{3} = \frac{0}{3} = 0
\]

2. **For \( n = 2 \):**

\[
a_2 = \frac{(-1)^{2+1} \cdot (2-1)}{2(2) + 1} = \frac{(-1)^3 \cdot 1}{5} = \frac{-1}{5} = -\frac{1}{5}
\]

3. **For \( n = 3 \):**

\[
a_3 = \frac{(-1)^{3+1} \cdot (3-1)}{2(3) + 1} = \frac{(-1)^4 \cdot 2}{7} = \frac{2}{7}
\]

4. **For \( n = 4 \):**

\[
a_4 = \frac{(-1)^{4+1} \cdot (4-1)}{2(4) + 1} = \frac{(-1)^5 \cdot 3}{9} = \frac{-3}{9} = -\frac{1}{3}
\]

5. **For \( n = 5 \):**

\[
a_5 = \frac{(-1)^{5+1} \cdot (5-1)}{2(5) + 1} = \frac{(-1)^6 \cdot 4}{11} = \frac{4}{11}
\]

### First Five Terms:
The first five terms of the sequence are:

\[
a_1 = 0, \quad a_2 = -\frac{1}{5}, \quad a_3 = \frac{2}{7}, \quad a_4 = -\frac{1}{3}, \quad a_5 = \frac{4}{11}
\]

Would you like further details or have any questions?

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**Related Questions:**

1. What is the general behavior of the sequence as \( n \) increases?
2. How does the sequence change if we modify the exponent of \((-1)^{n+1}\)?
3. What is the sum of the first \( n \) terms of this sequence?
4. Can the sequence be expressed as a sum of simpler sequences?
5. How would the sequence change if the numerator and denominator were swapped?

**Tip:** When analyzing sequences, it is helpful to look at the signs and magnitudes of terms separately to understand their pattern.

Discover the first five terms of the sequence a_n = (-1)^(n+1) * (n-1) / (2n+1). Detailed step-by-step calculation and analysis included. Suitable for advanced high school students.

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