Sums of odd numbers

$1 = 1^2$ . $1+3 = 4 = 2^2$ . $1+3+5 = 9 = 3^2$ . $1+3+5+7 = 16 = 4^2$ . Every time you add up the first few odd numbers starting from $1$ , you land exactly on a perfect square — and this is not a coincidence that happens to work for a few small cases and then quietly breaks down. The chapter A Square and A Cube first shows you this pattern as a number trick, something to notice and check by hand. The Algebra Play chapter then asks a sharper question: can you prove it holds for every whole number $n$ , not just the ones small enough to add up on paper? That is exactly the shift from arithmetic to algebra — replacing specific numbers with the letter $n$ and showing the pattern must hold no matter what $n$ is. This section builds that proof from a single idea: consecutive squares differ by consecutive odd numbers. Once you see why, the whole pattern falls out almost for free.

The core insight

The sum of the first $n$ odd numbers is always $n^2$ , because each new odd number added is exactly the gap between one perfect square and the next.

The core idea

Writing the $n$ -th odd number algebraically. Every odd number can be produced by the formula

2n - 1 \quad \text{for } n = 1, 2, 3, \dots

where: $n$ is the position of the odd number in the list ( $n=1$ gives the first odd number, $n=2$ the second, and so on). Check it: $n=1 \Rightarrow 1$ , $n=2 \Rightarrow 3$ , $n=3 \Rightarrow 5$ , $n=4 \Rightarrow 7$ — this correctly generates $1, 3, 5, 7, \dots$ in order, and it is the algebraic handle that makes the whole pattern provable rather than just observed.

The claim to prove. Adding up the first $n$ odd numbers, starting from $1$ , always gives a perfect square:

1 + 3 + 5 + \cdots + (2n-1) = n^2

where: the left side has exactly $n$ terms, the last of which is the $n$ -th odd number $2n-1$ , and the right side is that same $n$ squared. This is the pattern you can check for $n=1,2,3,4$ above; the rest of this section shows why it can never fail.

The key structural fact. Squares grow by steps, and each step is itself an odd number. Look at how $1^2, 2^2, 3^2, 4^2$ are built one on top of the other:

1^2 = 1, \qquad 2^2 = 1^2 + 3, \qquad 3^2 = 2^2 + 5, \qquad 4^2 = 3^2 + 7

Each square equals the previous square plus the next odd number. That single observation — proved algebraically below — is the entire engine behind the sum-of-odd-numbers pattern: adding odd numbers one at a time is exactly the same process as building squares one step at a time, so after $n$ steps you have added $n$ odd numbers and reached $n^2$ .

Figure

A staircase of four dot squares of side 1 2 3 and 4 with each new L shaped layer of dots highlighted and labelled with the odd number it adds

The formula, proved

Derivation AWhy the first n odd numbers add up to n squared

The goal is to show, using nothing but the distributive law of algebra, that adding the first $n$ odd numbers starting from $1$ always produces exactly $n^2$ , for every whole number $n$ and not merely for the small cases that can be checked by hand-addition — this is the difference between spotting a pattern and proving it.

Step 1. Look at the gap between two consecutive perfect squares, $(n+1)^2$ and $n^2$ . Expand $(n+1)^2$ using the distributive law: $(n+1)^2 = n^2 + 2n + 1$ . Subtracting $n^2$ from both sides gives the gap:

(n+1)^2 - n^2 = (n^2+2n+1) - n^2 = 2n+1

Step 2. Recognise $2n+1$ as an odd number. Compare it to the odd-number formula $2n-1$ : the expression $2n+1$ is simply the next odd number after $2n-1$ , i.e. the $(n+1)$ -th odd number. So Step 1 says: the gap between consecutive squares is always the next odd number in the list. Check it directly: $2^2-1^2=3$ , $3^2-2^2=5$ , $4^2-3^2=7$ — each gap matches the next odd number exactly.

Step 3. Chain the gaps together from $n=1$ up to build the whole sum. Since $1^2=1$ , and each further square is the previous square plus the next odd number, writing out the chain of gaps term by term gives

1^2 = 1,\quad 2^2 = 1^2+3,\quad 3^2=2^2+5,\quad \dots,\quad n^2 = (n-1)^2+(2n-1)

Adding all these steps together, every intermediate square cancels except the last one, leaving exactly the sum of odd numbers on one side and $n^2$ on the other:

1 + 3 + 5 + \cdots + (2n-1) = n^2

Step 4. Confirm the argument covers every $n$ , not just small ones. Nothing in Steps 1–3 depended on $n$ being small — the distributive-law expansion $(n+1)^2=n^2+2n+1$ is an algebraic identity true for every whole number, so the chain of gaps in Step 3 can be extended as far as you like. This is exactly what turns a pattern noticed for $n=1,2,3,4$ into a fact proved for all $n$ .

Worked example

Problem. Using the pattern, find $1+3+5+7+9+11+13$ without adding term by term, then verify by direct addition and identify why the running totals themselves form a familiar sequence.

Step 1. Count the terms. The list is $1,3,5,7,9,11,13$ ; using $2n-1=13$ gives $2n=14$ , so $n=7$ . These are the first $7$ odd numbers.

Step 2. Apply the pattern directly. Since the sum starts at $1$ and has $n=7$ terms,

1+3+5+7+9+11+13 = 7^2 = 49

Step 3. Verify by direct step-by-step addition: $1+3=4$ , $4+5=9$ , $9+7=16$ , $16+9=25$ , $25+11=36$ , $36+13=49$ . Both methods agree.

Step 4. Notice the running totals: $4, 9, 16, 25, 36, 49$ are themselves $2^2, 3^2, 4^2, 5^2, 6^2, 7^2$ — exactly the "each square is the previous square plus the next odd number" chain from Step 3 of the derivation, playing out one addition at a time.

The sum is $49$ .

Competitive stretch

A student claims $1+3+5+\cdots+19 = 100$ . The odd numbers from $1$ to $19$ are $1,3,5,7,9,11,13,15,17,19$ — ten terms, since $2n-1=19$ gives $n=10$ . The pattern gives $n^2=10^2=100$ , so the claim is correct. Pairing first-with-last confirms it independently: $(1+19)+(3+17)+(5+15)+(7+13)+(9+11) = 20+20+20+20+20 = 100$ , since there are $n/2 = 5$ pairs each summing to $2n=20$ , giving $\frac{n}{2}\times 2n = n^2$ — a second, purely arithmetic route to the same identity, though it needs a small adjustment when $n$ is odd and a middle term is left unpaired.

Common traps

Using $2n$ instead of $2n-1$ . The $n$ -th even number is $2n$ ; swapping it in for the $n$ -th odd number gives a completely different, wrong sum.
Miscounting the number of terms. "Odd numbers up to 19" does not mean $n=19$ — there are only $10$ odd numbers between $1$ and $19$ , so $n=10$ .
Applying $n^2$ to a sum that doesn't start at 1. A sum like $5+7+9+11$ is not covered by this shortcut; the identity only holds for odd numbers starting from $1$ .
Trusting a few checked cases as proof. Verifying $n=1,2,3$ is a good sanity check, but only the algebraic argument — via $(n+1)^2-n^2=2n+1$ — shows the pattern holds for every $n$ .

Ask the tutor: Curious how the pairing trick ( $\text{first}+\text{last}$ ) also lands on $n^2$ , or want to try the shortcut on a sum of your own choosing?

Why does adding odd numbers give a perfect square instead of some other kind of number

Because each odd number added is exactly the gap between one perfect square and the next one, as shown by $(n+1)^2-n^2=2n+1$ . Adding odd numbers one at a time is the same process, step for step, as building up squares $1^2, 2^2, 3^2,\dots$ one layer at a time. After adding $n$ odd numbers you have taken exactly $n$ such steps from $0$ , which lands you precisely on $n^2$ — it is not a coincidence but a direct consequence of how squares grow.

Does this pattern work if I start adding from an odd number other than 1, like 5+7+9+11

No, not with the plain $n^2$ formula — that shortcut is specifically for sums that begin at $1$ . A sum like $5+7+9+11$ can still be handled algebraically (for example by writing it as the full sum up to $11$ minus the sum up to $3$ , i.e. $6^2 - 2^2$ ), but you cannot just count the four terms and square them. Always check where the sum starts before reaching for $n^2$ .

How is this different from just checking the pattern with a calculator for a few values of n

Checking $n=1,2,3,4$ tells you the pattern is true for those specific cases, but it cannot tell you it is true for every whole number — there could always be some larger $n$ where it breaks. The algebraic proof in this section uses the distributive law to show $(n+1)^2-n^2=2n+1$ for any $n$ , and chains that fact together to reconstruct the full sum. That is what turns "true so far" into "true always," which is the whole point of the Algebra Play chapter.