Area preservingDifficulty · Easy

Odd Numbers Sum to a Square (1+3+5+…=n²)

Add the odd numbers 1, 3, 5, 7, … and you always get a perfect square. Wrapping L-shaped gnomons around a square shows why.

Problem

Show that the sum of the first n odd numbers 1+3+5+…+(2n−1) equals n², by growing a square.

Cut · rotate · rearrange

Press Play or move Next one step at a time. Follow the pieces until the answer becomes visible. (Arrow keys and Space also work.)

135791+3+5+7+9 = 25 = 5²

Start with a 1×1 square (1 = 1²).

Why this approach works

Instead of summing, grow a square layer by layer. Start with 1×1 (=1²). Wrap an L-shaped gnomon to reach 2×2 — that adds 3 cells. The next gnomon adds 5, then 7, …; growing k×k to (k+1)×(k+1) always adds 2k+1 cells, an odd number. So adding 1+3+5+… is the same as growing the square, and after n terms the n×n square is full: 1+3+5+…+(2n−1) = n².

Proof

Area = 1 + 3 + 5 + … + (2n−1) = n²

Growing a k×k square to (k+1)×(k+1) adds a gnomon of 2k+1 cells (odd). Summing the odd numbers is thus growing the square, and after n terms it is n×n.

Try it yourself

Count each gnomon: 1, 3, 5, 7, 9 — all odd. Together they fill the 5×5 = 25 square.

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