Sum of Odd Numbers - Formula, Proof & Examples

The sum of the first $n$ odd numbers is $n^2$. Written as a formula: $$1 + 3 + 5 + \dots + (2n-1) = n^2$$ Here $(2n-1)$ is the $n$-th odd number: the 1st is $2(1)-1 = 1$, the 4th is $2(4)-1 = 7$, and so on. An odd number is any integer not divisible by 2, so the odd numbers are $1, 3, 5, 7, 9, \dots$, each two more than the one before.

Because each term is 2 more than the last, the odd numbers form an arithmetic progression with first term 1 and common difference 2 — which is why a tidy closed formula exists at all.

Sum of Odd Numbers Formula

The sum of the first $n$ odd numbers is: $$1 + 3 + 5 + \dots + (2n-1) = n^2$$

The formula can also be reached through the arithmetic-progression sum $S_n = \tfrac{n}{2}(a + l)$, with first term $a = 1$ and last term $l = 2n-1$: $$S_n = \frac{n}{2}\big(1 + (2n-1)\big) = \frac{n}{2}(2n) = n^2$$

Examples of the Sum of Odd Numbers

Example 1

Find the sum of the first 5 odd numbers.

The first 5 odd numbers are $1, 3, 5, 7, 9$. Here $n = 5$. $$\text{Sum} = n^2 = 5^2 = 25$$ Check by adding: $1 + 3 + 5 + 7 + 9 = 25$. ✓

Example 2

Find the sum of the first 10 odd numbers.

The tempting move is to average the first and last term and multiply by 10, the way you might for any list. First find the 10th odd number: $2(10)-1 = 19$. $$\text{Sum} = \frac{1 + 19}{2} \times 10 = 10 \times 10 = 100$$ That happens to give 100, and it is correct, but it hides the cleaner fact. The direct rule is faster and shows why. $$\text{Sum} = n^2 = 10^2 = 100$$ The rescue: for consecutive odd numbers starting at 1, skip the average-of-endpoints setup and use $n^2$ directly.

Example 3

What is the sum $1 + 3 + 5 + \dots + 99$?

The last term is 99, so solve $2n - 1 = 99$. $2n = 100$ $n = 50$ $$\text{Sum} = 50^2 = 2500$$

Example 4

The sum of the first $n$ odd numbers is 64. Find $n$.

$n^2 = 64$ $n = \sqrt{64} = 8$ There are 8 odd numbers, namely $1, 3, 5, 7, 9, 11, 13, 15$.

Example 5

Find the sum of odd numbers from 1 to 15.

The odd numbers from 1 to 15 are $1, 3, 5, 7, 9, 11, 13, 15$, that is 8 terms. $$\text{Sum} = 8^2 = 64$$

Example 6

Find the sum of the first 20 odd numbers, then subtract the sum of the first 12.

Sum of first 20: $20^2 = 400$ Sum of first 12: $12^2 = 144$ Difference: $400 - 144 = 256$ The sum of the 13th through 20th odd numbers is 256.

Why the Sum of Odd Numbers Builds a Square

The formula is not a lucky accident — it is a picture. Start with one dot, a $1 \times 1$ square. To grow it into a $2 \times 2$ square you add an L-shaped layer along one side and the bottom: that L holds exactly 3 dots. To reach $3 \times 3$ you add an L of 5 dots; to reach $4 \times 4$, an L of 7.

• Each L-shaped layer holds the next odd number of dots.

• Each layer completes the next perfect square.

• So $1 + 3 + 5 + \dots + (2n-1)$ is the $n \times n$ square, dot for dot.

This visual is why the Pythagoreans called these figurate numbers "gnomons," and it links directly to square numbers: every perfect square is a running total of odd numbers. The Pythagoreans studied this around 500 BCE, long before algebraic notation existed — the geometry came first.

The same additive-layer idea underlies how the sum of the first n natural numbers gets its triangular-number formula; odd numbers give squares, all naturals give triangles.

Properties of the Sum of Odd Numbers

A few properties follow directly from the $n^2$ rule and are worth keeping in mind:

• The sum is always a perfect square. The sum of the first $n$ odd numbers is exactly $n^2$, so it is never irrational or fractional.

• The odd numbers form an arithmetic progression. First term $1$, common difference $2$, $n$-th term $2n-1$.

• The mean of the first $n$ odd numbers is $n$. Since the total is $n^2$ spread over $n$ terms, the average is $n^2 / n = n$.

• Runs that do not start at 1 need adjusting. To add odd numbers that start above 1, take the sum up to the last term and subtract the sum of the odds you skipped. For example, $5 + 7 + 9 = (1+3+5+7+9) - (1+3) = 25 - 4 = 21$.

Tripping Points to Avoid When Calculating the Sum of Odd Numbers

Mistake 1: Using $n^2$ when the count is wrong

Where it slips in: Reading "sum up to 99" and plugging in $n = 99$.

Don't do this: Compute $99^2$ when 99 is the last term, not the count of terms.

The correct way: First find how many odd numbers there are: solve $2n - 1 = 99$ to get $n = 50$, then square. The habit that fixes this is always converting a last-term to a term-count before applying $n^2$.

Mistake 2: Including even numbers by accident

Where it slips in: Listing "1, 2, 3, 4, 5" and calling it the first five odd numbers.

Don't do this: Treat consecutive integers as consecutive odd numbers.

The correct way: Odd numbers step by 2: $1, 3, 5, 7, 9$. Every term must be odd, so the count and the last term both differ from the natural-number list.

Mistake 3: Assuming the rule works for odd numbers not starting at 1

Where it slips in: Applying $n^2$ to $3 + 5 + 7$.

Don't do this: Claim $3 + 5 + 7 = 3^2 = 9$.

The correct way: The $n^2$ rule needs the run to start at 1. For $3 + 5 + 7$, take the sum up to 7 ($4^2 = 16$) minus the first term ($1$): $16 - 1 = 15$. The confusion between "first $n$ odd numbers" and "any three odd numbers" is where most wrong answers come from.

Conclusion

• The sum of odd numbers starting at 1 is always a perfect square: $1 + 3 + \dots + (2n-1) = n^2$.

• The $n$-th odd number is $2n - 1$, so convert a last-term to a term-count before squaring.

• The visual proof builds an $n \times n$ square from L-shaped layers of $1, 3, 5, \dots$ dots.

• The rule needs the run to start at 1; otherwise subtract the missing head of the sequence.

To take the sum of odd numbers further with a teacher, work with a Bhanzu algebra tutor, join algebra classes, or browse math tutoring. For a live walkthrough, book a free demo class.

Read More

• Sum of Cubes of n Natural Numbers — another figurate-number pattern.

• Squares and Square Roots — the squares this sum lands on.

• Square Root 1 to 25 — the roots of those perfect squares.

• Polynomials — where these sum formulas generalise.