How to Prove the Sum of the First n Odd Numbers by Induction

The first $n$ positive odd numbers are $1,3,5,\ldots,2n-1$. Their sum has a simple closed form: it is $n^2$. Mathematical induction proves this by showing that the formula works for the first value and that its truth for one value forces it to hold for the next.

State the claim precisely

For every positive integer $n$, define

$$P(n):\quad \sum_{i=1}^{n}(2i-1)=n^2.$$

The summand $2i-1$ is the $i$th positive odd number: when $i=1$ it is $1$, when $i=2$ it is $3$, and so on. Thus the left-hand side adds the first $n$ odd numbers.

Base case: $n=1$

When $n=1$, the sum contains just one term:

$$\sum_{i=1}^{1}(2i-1)=2(1)-1=1.$$

The right-hand side is also $1^2=1$. Therefore, $P(1)$ is true.

Inductive step

Assume the statement is true for an arbitrary positive integer $k$. This assumption is the inductive hypothesis:

$$\sum_{i=1}^{k}(2i-1)=k^2.$$

To prove the next case, consider the sum through $k+1$. It consists of the sum through $k$, together with one new term, the $(k+1)$st odd number:

$$\begin{aligned} \sum_{i=1}^{k+1}(2i-1) &=\sum_{i=1}^{k}(2i-1)+\bigl(2(k+1)-1\bigr)\\ &=k^2+(2k+1)\\ &=k^2+2k+1\\ &=(k+1)^2. \end{aligned}$$

The second line uses the inductive hypothesis. The final expression is exactly the right-hand side of $P(k+1)$. Therefore, if $P(k)$ is true, then $P(k+1)$ is true.

Conclusion by induction

The statement holds for $n=1$, and its truth for any positive integer $k$ implies its truth for $k+1$. By mathematical induction,

$$\boxed{\sum_{i=1}^{n}(2i-1)=n^2\quad\text{for every positive integer }n.}$$

Check with a concrete value

Take $n=4$. The sum is $1+3+5+7=16$, and the formula gives $4^2=16$. In the inductive step, this same calculation appears as extending the sum through $3$:

$$1+3+5=3^2=9,\qquad 9+7=9+(2\cdot3+1)=16=4^2.$$

The example illustrates the induction step: once the formula is known through $k$, adding the next odd number changes $k^2$ into $(k+1)^2$.

Why adding the next odd number works

The difference between consecutive squares is

$$(k+1)^2-k^2=2k+1.$$

That difference is precisely the next odd number after $1,3,\ldots,2k-1$. So the sum grows from $k^2$ to $(k+1)^2$ when that odd number is added. This observation explains why the formula fits the pattern; the induction proof makes the reasoning apply to every positive integer, not just to a few examples.

Common mistakes

• Checking only a few values. Examples such as $n=1,2,3$ can suggest the formula, but they do not prove it for all positive integers. The inductive step is what covers every subsequent value.
• Adding the wrong next term. After the sum through $k$, the next index is $k+1$, so its term is $2(k+1)-1=2k+1$, not $2k-1$.
• Assuming what you need to prove. In the inductive step, assume only the statement for $k$. Then use that assumption to derive the statement for $k+1$; do not assume the formula already holds for all $n$.
• Forgetting the base case. The step shows how truth passes from one integer to the next, but a starting case is needed to begin that chain.

If the index range is taken to include $n=0$, the empty sum is defined to be $0$, and the formula also holds there because $0^2=0$. The usual induction proof above starts at $n=1$, the first positive integer.