The \(n\)th term is \(a_n = S_n - S_{n-1}\).

["# Understanding the nth Term: (a_n = S_n - S_{n-1})", "In mathematics, especially in the study of sequences and series, one of the most powerful and elegant relationships lies at the heart of understanding how sequences evolve. The formula (a_n = S_n - S_{n-1}) unlocks a foundational insight: the (n)th term of a sequence is simply the difference between the sum of the first (n) terms and the sum of the first ((n-1)) terms.", "## What Are (S_n) and (S_{n-1})?", "Let (S_n) denote the (n)th partial sum of a sequence ( {a_n} ). That is:", "[\nS_n = a_1 + a_2 + a_3 + \dots + a_{n}\n]", "Similarly, (S_{n-1}) represents the sum up to the ((n-1))th term:", "[\nS_{n-1} = a_1 + a_2 + \dots + a_{n-1}\n]", "By definition, (S_n) includes all terms from (a_1) to (a_n), while (S_{n-1}) stops one term short—allowing us to extract the very next term, (a_n), with a simple subtraction.", "## The Simplifying Insight: (a_n = S_n - S_{n-1})", "From the definitions above, it follows directly that:", "[\na_n = (\underbrace{a_1 + a_2 + \dots + a_{n-1} + a_n}) - (\underbrace{a_1 + a_2 + \dots + a_{n-1}})\n]", "The two substrings cancel out, leaving:", "[\na_n = a_n\n]", "This confirms the validity of the formula. But far more importantly, it reveals a profound truth: each term in a sequence is derived by subtracting a "partial sum" from the next, encapsulating both accumulation and increment.", "## Why This Formula Matters", "### 1. Connecting Sequences and Series\nThe formula bridges discrete sequences and continuous series. It shows that series are essentially cumulative processes—each new term builds on the sum before it.", "### 2. Recursive Sequences Made Simple\nIn recursive definitions, knowing (S_n) allows instant computation of (a_n), eliminating the need for iterative addition unless working iteratively.", "### 3. Foundation for Convergence Tests\nIn analysis, examining (S_n - S_{n-1} = a_n) helps determine if a sequence converges. If (a_n \ o 0) as (n \ o \infty), the series may converge—this link begins here.", "### 4. Algorithmic Efficiency\nIf the sum (S_n) is efficiently computable (e.g., via closed forms), this formula provides a direct way to compute terms without nested summations.", "## Practical Examples", "### Example 1: Arithmetic Sequence\nLet (a_n = 3n). Then:\n[\nS_n = \frac{n(1 + 3n)}{2} = \frac{n + 3n^2}{2}, \quad S_{n-1} = \frac{(n-1) + 3(n-1)^2}{2}\n]\nCompute the difference:\n[\na_n = S_n - S_{n-1} = \frac{n + 3n^2 - (n-1) - 3(n^2 - 2n + 1)}{2} = 3n - (3n - 3) = 3\n]\nConfirmed: each term is (3).", "### Example 2: Geometric Sequence\nLet (a_n = 2^n).\n[\nS_n = 2 + 2^2 + \dots + 2^n = 2(2^n - 1), \quad S_{n-1} = 2(2^{n-1} - 1)\n]\nThen:\n[\na_n = S_n - S_{n-1} = 2(2^n - 1) - 2(2^{n-1} - 1) = 2^{n+1} - 2 - 2^n + 2 = 2^n\n]\nAgain, correct.", "## Key Takeaways", "- (a_n = S_n - S_{n-1}) is not just a formula—it’s a conceptual breakthrough.\n- It reveals how sequences build incrementally, each term growing from prior sums.\n- This relationship is pivotal in proving convergence, analyzing recursive definitions, and optimizing computational methods.", "Mastering this relationship equips learners and researchers alike with a deeper lens into the structure of mathematical sequences. Whether studying advanced calculus, numerical analysis, or theoretical computer science, understanding (a_n = S_n - S_{n-1}) is essential.", "Begin applying this insight next time you study sequences—unlock the hidden elegance of summation and difference!"]