Sum of 1 to n natural numbers explained

You need the sum of 1 to n, say 1 to 100, and you need it right now. The formula n(n+1)/2 gives you the answer in one step. 100 times 101 divided by 2 equals 5050. No adding, no spreadsheet, no risk of an off-by-one error. Here is the formula, why it works, and how to handle sums that do not start at 1, plus sums of even and odd numbers. The sum of 1 to n is the foundation for every other finite sum you will meet in precalculus, calculus, and discrete maths.

The sum of the first n natural numbers is S = n(n+1)/2. For n = 10, S = 10×11/2 = 55. For n = 50, S = 50×51/2 = 1275. The formula works for any positive integer n. It is a polynomial of degree 2, which means the sum grows roughly as n²/2.

The closed form comes from the arithmetic series formula S = n/2 × (first + last). Here the first term is 1, the last term is n, and the number of terms is n, so S = n/2 × (1 + n) = n(n+1)/2.

The formula is often taught through an anecdote about Karl Friedrich Gauss. In 1784 or 1785, at age 9 or 10, Gauss allegedly summed 1 to 100 in seconds by pairing 1+100, 2+99, 3+98, and so on, getting 50 pairs each summing to 101, for a total of 5050.

The story first appears in a biography by Sartorius von Waltershausen in 1856, more than 60 years after the event. Brian Hayes, writing in American Scientist (May, June 2006, Volume 94, Number 3, pages 16-21), calls the anecdote "almost certainly apocryphal." No contemporary source confirms it. The pairing method itself is valid for any arithmetic series with an even number of terms. Whether Gauss used it is historically uncertain, but the mathematics is sound.

When the sum of consecutive integers does not start at 1, subtract the sum of the unwanted lower part. To sum from a to b inclusive, compute S(1 to b) minus S(1 to a−1).

S(a to b) = b(b+1)/2 − (a−1)a/2.

For example, sum 50 to 100. S(1 to 100) = 5050. S(1 to 49) = 49×50/2 = 1225. Difference = 5050 − 1225 = 3825.

If the index start value is not 1, the number of terms is (b − a + 1), not b. Misidentifying the count is a common failure mode. The upper bound and the number of terms are the same only when the lower bound is 1.

The first n even numbers are 2, 4, 6, ..., 2n. Their sum is 2 times the sum of 1 to n: 2 × n(n+1)/2 = n(n+1).

Sum of first 10 even numbers: 10×11 = 110. Check: 2+4+6+8+10+12+14+16+18+20 = 110.

The sum of even numbers formula grows linearly with n², just as the original sum does, but without the division by 2.

The first n odd numbers are 1, 3, 5, ..., (2n−1). Their sum is n².

Sum of first 10 odd numbers: 10² = 100. Check: 1+3+5+7+9+11+13+15+17+19 = 100.

This is a special case of the arithmetic series formula with common difference 2. The sum of odd numbers formula is the simplest closed form of them all: no coefficient, no division, just the square of the count.

For the sum of consecutive integers 1 to n:

n = 10: 55
n = 50: 1275
n = 100: 5050

These values come from n(n+1)/2. Use them as a sanity check when solving textbook problems. If your manual calculation gives a different result for n=100, you have an error, likely an off-by-one in the upper bound or a misapplied constant multiple property.

A short induction proof confirms the formula holds for every n.

Base case: n=1. Left side: 1. Right side: 1(1+1)/2 = 1. True.

Inductive step: Assume S(k) = k(k+1)/2 for some k≥1. Show S(k+1) = (k+1)(k+2)/2.

S(k+1) = S(k) + (k+1) = k(k+1)/2 + (k+1) = (k+1)[k/2 + 1] = (k+1)(k+2)/2.

The induction proof is standard in discrete-maths textbooks and verifies the closed form without relying on the pairing diagram or the Gauss anecdote.

• Off-by-one in the upper bound: Writing Σ_{k=1}^{n} when the last term should be n−1 adds an extra term. Always check the last term against the problem statement.
• Confusing number of terms with upper bound: For Σ_{k=5}^{10} k, the upper bound is 10, but the number of terms is 6. Using n(n+1)/2 directly gives the wrong result.
• Misapplying the formula for sums that do not start at 1: Subtract S(1 to a−1) from S(1 to b). Do not try to adapt n(n+1)/2 by shifting the index without adjusting bounds.

Compute the sum of 1 to 100 using n(n+1)/2. You get 5050. Write it down. Then check it against the pairing method: 50 pairs of 101, as in the Gauss story. If both match, your understanding is solid. If they do not, recheck the formula and the number of terms. The single thing that most often goes wrong here is treating the upper bound as the number of terms when the lower bound is not 1. Always count the terms explicitly.