Sum of Natural Numbers Formula — Proof & Examples

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Sum of Natural Numbers Formula — Proof & Examples

TL;DR

The sum of natural numbers formula is ∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}∑k=1nk=2n(n+1) — the cleanest closed-form result in elementary mathematics, discovered by an eight-year-old in his school classroom. This article gives the formula, two proofs (pairing and induction), three worked examples spanning Quick to Stretch, the related sum-of-squares and sum-of-cubes identities, and the most common slip-ups.

A Story That Begins With a Bored Schoolboy

When his teacher told the class to add the numbers 111 to 100100100 to keep them busy, eight-year-old Carl Friedrich Gauss had the answer in under a minute.

The sum of natural numbers formula — also called the Gauss sum or the triangular number formula — is:

∑k=1nk=1+2+3+…+n=n(n+1)2.\sum_{k=1}^{n} k = 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}.

For n=100n = 100, that gives 100⋅1012=5050\frac{100 \cdot 101}{2} = 50502100⋅101=5050 — the answer the young Gauss called out before the chalk dust had settled.

The Formula

For any positive integer n:

;Sn=∑k=1nk=1+2+3+…+n=n(n+1)2;\boxed{;S_n = \sum_{k=1}^{n} k = 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2};};

The companion identities:

∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \quad\text{(sum of first $n$ squares)}

∑k=1nk3=(n(n+1)2)2\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \quad\text{(sum of first $n$ cubes — equals the square of the sum of first $n$ integers)}

The cube-sum identity is the most elegant: ∑k3=(∑k)2\sum k^3 = (\sum k)^2.

Quick facts.

Closed form: Sn=n(n+1)2S_n = \frac{n(n+1)}{2}.

Also called: the nnnth triangular number TnT_nTn.

Grade introduced: CCSS-M HSA-SSE.B.4; NCERT Class 10 Chapter 5 — Arithmetic Progressions.

First few values: T1=1,T2=3,T3=6,T4=10,T5=15,T6=21,T7=28,T8=36,T9=45,T10=55T_1 = 1, T_2 = 3, T_3 = 6, T_4 = 10, T_5 = 15, T_6 = 21, T_7 = 28, T_8 = 36, T_9 = 45, T_{10} = 55.

Special case of AP sum: Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d] with a=1,d=1a = 1, d = 1.

Two Proofs

Proof 1 — Pairing (Gauss's method).

Write the sum forwards and backwards on two lines:

S=1+2+3+…+(n−1)+nS = 1 + 2 + 3 + \ldots + (n-1) + n

S=n+(n−1)+(n−2)+…+2+1S = n + (n-1) + (n-2) + \ldots + 2 + 1

Add term-by-term. Every column sums to n+1n + 1. There are nnn columns, so:

2S=n(n+1).

Divide by 222: S=n(n+1)2S = \frac{n(n+1)}{2}.

Proof 2 — Induction.

Base case (n=1): S1=1=1⋅22=1S_1 = 1 = \frac{1 \cdot 2}{2} = 1.

Inductive step. Assume Sn=n(n+1)2S_n = \frac{n(n+1)}{2}. Then:

Sn+1=Sn+(n+1)=n(n+1)2+(n+1)=n(n+1)+2(n+1)2=(n+1)(n+2)2.S_{n+1} = S_n + (n+1) = \frac{n(n+1)}{2} + (n+1) = \frac{n(n+1) + 2(n+1)}{2} = \frac{(n+1)(n+2)}{2}.

This is the formula with n replaced by n+1. The pattern propagates. By induction, Sn=n(n+1)2S_n = \frac{n(n+1)}{2} for all positive integers.

The pairing proof is the one a child can grasp. The induction proof is what an algebraist would write.

Three Worked Examples, From Quick to Stretch

Quick. Find 1+2+3+…+101 + 2 + 3 + \ldots + 101+2+3+…+10.

S10=10⋅112=55.S_{10} = \frac{10 \cdot 11}{2} = 55.

Final answer: 555555.

Standard (Wrong-Path-First). Find 50+51+52+…+100.

Wrong path. A student in our McKinney TX Grade 9 cohort once wrote: "S100=100⋅1012=5050". That's the sum from 111 to 100, not from 50 to 100. The student substituted into the formula without checking that the formula assumes the sum starts at 1.

Correct. Use subtraction:

∑k=50100k=∑k=1100k−∑k=149k=100⋅1012−49⋅502=5050−1225=3825.\sum_{k=50}^{100} k = \sum_{k=1}^{100} k - \sum_{k=1}^{49} k = \frac{100 \cdot 101}{2} - \frac{49 \cdot 50}{2} = 5050 - 1225 = 3825.

Final answer: 50+51+…+100=3825.

Stretch. Find n such that the sum 1+2+3+…+n=210.

Set n(n+1)2=210. Then n2+n=420n^2 + n = 420, so n^2+n−420=0.

Quadratic formula: n=−1±1+16802=−1±16812=−1±412.

Positive root: n=20.

Final answer: n=20. Check: 20⋅212=210.

Where the Formula Lives — Beyond Counting

The sum-of-naturals formula isn't a one-off classroom trick — it sits at the foundation of several mathematical structures.

Handshakes in a room. The number of distinct handshakes between n+1 people is (n+12)=n(n+1)2\binom{n+1}{2} = \frac{n(n+1)}{2} — exactly the sum-of-naturals.
Algorithm complexity. Bubble sort and insertion sort perform n(n−1)2 comparisons in the worst case — a Gauss-sum cousin one position offset.
Bowling-pin arrangements. A standard 10-pin bowling rack is T4=4⋅52=10T_4 = \frac{4 \cdot 5}{2} = 10 pins.
Stacking — pyramids and balls. The sum of the first n triangular numbers gives the tetrahedral numbers Tn(3)=n(n+1)(n+2)6T_n^{(3)} = \frac{n(n+1)(n+2)}{6} — the number of cannonballs you can stack in a triangular pyramid n layers high.

Pitfalls to Avoid With the Gauss Sum

1. Forgetting the formula assumes the sum starts at 1.

Where it slips in: Problems phrased "sum of natural numbers from 50 to 100."

The correct way: For sums that don't start at 1, compute the cumulative sum and subtract: ∑k=abk=b(b+1)2−(a−1)a2.

2. Mixing n with the count of terms when the sum doesn't start at 1.

Where it slips in: Same family of problems as Mistake 1.

The correct way: For a sum from a to b, the count of terms is b−a+1, not b.

3. Using the formula on non-integer "natural numbers."

Where it slips in: Word problems where the variable could be fractional.

The correct way: The formula assumes n is a positive integer.

4. Mixing the Gauss sum with the sum-of-squares formula.

Where it slips in: Problems that switch between ∑k, ∑k2, ∑k3 in succession.

The correct way: Each formula is its own object.

A Story About Gauss and the Schoolmaster

In 1786, when Carl Friedrich Gauss (1777–1855, Germany) was eight or nine years old, his arithmetic teacher in Brunswick — a man named Büttner — set the class to add the numbers 1 to 100. The exercise was intended to keep the class occupied for an hour while Büttner attended to other matters.

Gauss put his slate on the teacher's desk within minutes. "Ligget se," he said in his local dialect — "There it lies." On the slate was a single number: 5050.

He had paired 1+100=101, 2+99=101, 3+98=101, and so on — fifty pairs, each summing to 101. 50⋅101=5050.

Büttner was reportedly stunned. He sent away to Hamburg for the most advanced arithmetic book he could find, gave it to the boy, and Gauss soon outgrew his teacher's ability to teach him.

By age 19, Gauss had proven the constructibility of the regular 17-gon — a problem unsolved for 2000 years. By age 24, he had published Disquisitiones Arithmeticae and become Europe's leading mathematician.

The pairing trick is now the standard derivation of the sum-of-naturals formula taught to every student worldwide.

Conclusion

The sum of natural numbers formula for the first n positive integers is Sn=n(n+1)2.
The pairing proof (Gauss's method) is short — write the sum twice, once forwards and once backwards, and add.
The sum of the first n cubes equals the square of the sum: ∑k3=(∑k)2 — a uniquely clean identity.
The most common mistake is forgetting that the formula counts from 1 — always check the lower bound first.

Quick Self-Check — Try These

Try these before moving on. If you slip on starting-point, come back to Mistake 1.

Find 1+2+3+…+50 + 2 + 3 + … + 50.
Find 30+31+32+…+80.
For what n does n(n+1)2=1275? (Hint: solve a quadratic.)

✍️ Written By

Content Creator and Editor

Bhanzu’s editorial team, known as Team Bhanzu, is made up of experienced educators, curriculum experts, content strategists, and fact-checkers dedicated to making math simple and engaging for learners worldwide.