B.5.RA:5.1 - Understanding why the sum of odd numbers is a square

Consider this compressed argument: “The sum of the first n positive odd numbers is n²: the sum and the square start at zero, and both increase by 2n+1 when n increases by one.” The statement concerns every nonnegative integer n. A reader recognizes the formula but needs to explain the steps compressed in that reason.

Write S(0)=0 and S(n+1)=S(n)+(2n+1). The main reason is that both the sum and the square start at zero and grow by the same amount when n increases by one. The algebraic identity (n+1)²−n²=2n+1 supplies that connection.

Recover the general transition. Assuming S(n)=n² for an arbitrary nonnegative integer n gives:

S(n+1)=S(n)+2n+1=n²+2n+1=(n+1)².

The temporary assumption is the induction hypothesis. It supports the successor step. Together with S(0)=0, that step establishes the statement for every nonnegative integer by induction.

For n=3, the sum is 1+3+5=9. Adding the next odd number, 7, gives 16. This instance makes the equal-increment operation visible. The argument’s reach comes from the arbitrary n, the base value and the induction rule.

A square drawing gives another way to follow the increment: grow an n-by-n square with a row of n cells and a column of n+1 cells. That adds 2n+1 cells. The drawing and algebra expose the same increment under the counting interpretation.

The reader can now explain the role of the initial value and successor step. If the next task instead asks for the sum of n odd terms beginning at 3, the changed range opens a revision: use the established sum through the (n+1)th odd number and remove the first term, giving S(n+1)−1=n²+2n. For four terms, 3+5+7+9=24. The reusable contribution is the recovered relation between range, initial value and increment.

If the original question asked only for 1+3+5, direct addition would already supply the result. Recovering the general argument earns its effort when explanation, general use or revision needs it.