find the \( n \) the term of the Series \( 2,-5,8,-11,14, \ldots \)

We are given the series: 2, -5, 8, -11, 14, … Step 1. Look at the absolute values of the terms: 2, 5, 8, 11, 14, … Notice that these form an arithmetic sequence with a first term of 2 and a common difference of 3. Thus, the nth term of the absolute values can be written as:  Absolute term = 2 + (n - 1) · 3 = 3n - 1. Step 2. Determine the signs: Observe that the first term is positive, the second term is negative, the third is positive, etc. This indicates that the sign alternates with each term. A common way to represent this alternating sign is to use (-1) raised to a power that depends on n. Here, using (-1)^(n+1) will yield a positive sign for odd n and a negative sign for even n. Step 3. Combine the information: Thus, the nth term of the series is:  aₙ = (-1)^(n+1)·(3n - 1). This formula satisfies all given terms:  For n = 1: a₁ = (-1)^(1+1)·(3·1 - 1) = 1·(2) = 2  For n = 2: a₂ = (-1)^(2+1)·(3·2 - 1) = (-1)·(5) = -5  For n = 3: a₃ = (-1)^(3+1)·(3·3 - 1) = 1·(8) = 8 Therefore, the nth term of the series is: aₙ = (-1)^(n+1)(3n - 1).