Arithmetic Series (Gauss Sum Formula)

According to the anecdote, nine-year-old Gauss's class was told to add the numbers 1 to 100, and he had the answer within seconds: he paired the first with the last term, the second with the second-to-last and so on: $1 + 100 = 101$, $2 + 99 = 101$, $3 + 98 = 101$, ... That gives 50 pairs each summing to 101, hence $50\cdot{}101 = 5050$. Exactly this pairing sits inside the general formula s_n = n·(a₁ + a_n)/2: n terms form $\frac{n}{2}$ pairs with the constant sum a₁ + a_n. The trick works for every arithmetic series, because the growth at the front and the shrinking at the back balance exactly.

There are two equivalent forms. If you know the first AND the last term, take the pair form s_n = n·(a₁ + a_n)/2; it is the fastest. If instead you know a₁, the constant difference d and the count n, take s_n = n/2·(2a₁ + (n−1)·d), or first compute a_n = a₁ + (n−1)·d and then switch to the pair form. Example: $a_{1} = 5$, $d = 3$, $n = 20$: a_n = 5 + 19·3 = 62 and $s_{20} =\frac{ 20\cdot{}(5 + 62)}{2} = 670$; the d-form directly gives $10\cdot{}(10 + 57) = 670$ ✓. Both formulas are the same statement, once expressed with a_n and once with d.

With n = (a_n − a₁)/d + 1: range divided by step width, plus 1. The "+1" is crucial and most often forgotten, because the starting term counts too. Examples: from 1 to 100 in steps of one there are $\frac{(100 - 1)}{1} + 1 = 100$ terms. The even numbers from 2 to 100 have $d = 2$, so $\frac{(100 - 2)}{2} + 1 = 50$ terms, and their sum is $\frac{50\cdot{}(2 + 100)}{2} = 2550$. From 17 to 71 in steps of three: $\frac{(71 - 17)}{3} + 1 = 19$ terms. Check: if the division does not come out evenly, the supposed final term does not belong to the sequence at all; then determine the largest fitting term first.

Both are arithmetic series with $d = 2$, and two neat short formulas arise. The first n even numbers: 2 + 4 + ... $+ 2n = n\cdot{}(n + 1)$; example $n = 50$: $50\cdot{}51 = 2550$, identical to the pair formula $\frac{50\cdot{}(2 + 100)}{2}$ ✓. The first n odd numbers: 1 + 3 + ... $+ (2n - 1) = n^{2}$; example: $1 + 3 + 5 + 7 = 16 = 4^{2}$. The second formula has a famous geometric interpretation: laying an L-shaped border of the next odd number of stones around a square grows it square by square. Both identities are standard exercises for mathematical induction and quick check values in exams.

The building law of the terms. Arithmetic means: constant DIFFERENCE, each term arises by adding d (5, 8, 11, 14 with d = 3); the sum grows polynomially, formula s_n = n·(a₁ + a_n)/2. Geometric means: constant FACTOR, each term arises by multiplying by q (3, 6, 12, 24 with q = 2); for $|q| > 1$ the sum grows exponentially, formula s_n = a₁·(qⁿ − 1)/(q − 1). Quick test on data: differences of neighbouring terms constant → arithmetic; ratios constant → geometric. Typical applications: linear instalments and seat rows arithmetic, compound interest and growth processes geometric. Only the geometric series can have a finite limit, for $|q| < 1$.