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

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We begin by examining the series:
  2, -5, 8, -11, 14, …

Step 1. Notice that if we ignore the signs, the magnitudes form:
  2, 5, 8, 11, 14, … which is an arithmetic sequence with first term 2 and common difference 3. Thus, the nth term’s absolute value is:
  2 + (n - 1) · 3 = 3n - 1

Step 2. Next, observe the pattern of signs:
  n = 1 → positive,
  n = 2 → negative,
  n = 3 → positive, etc.
This alternation can be represented using (–1) raised to a power. A convenient choice is:
  (–1)^(n+1)
Because when n = 1, (–1)^(2) = 1 (positive), and when n = 2, (–1)^(3) = –1 (negative), which fits the series.

Step 3. Putting it all together, the general (nth) term is:
  aₙ = (–1)^(n+1) (3n – 1)

Thus, the nth term of the series is:
  aₙ = (–1)^(n+1)(3n – 1)
The nth term of the series is $$ a_n = (-1)^{n+1} (3n - 1) $$.

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

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