I’m practicing square numbers and noticed this pattern: 1 = 1, 1+3 = 4, 1+3+5 = 9, 1+3+5+7 = 16, etc. People say the sum of the first n odd numbers is always n^2. I can see it works for small cases, but I want to understand why it’s always true, not just memorize it.
Here’s my partial attempt: I tried induction. Base case n=1 is fine. Then I assumed 1+3+5+…+(2n−1) = n^2. If I add the next odd number, I get n^2 + (2n+1), which looks like (n+1)^2 – but I feel like I’m just manipulating symbols without seeing the reason. I’m also getting confused with the indexing: is the k-th odd number 2k−1 or 2k+1, and why does the “next odd” after 2n−1 come out as 2n+1 rather than 2n+1 being two steps ahead?
I also tried a picture: build an n×n square and then grow it to (n+1)×(n+1) by adding an L-shaped border. I keep second-guessing how many unit squares are in that L (and whether I’m double-counting the corner).
Could someone explain, step by step, why the sum of the first n odd numbers is exactly n^2, and how to keep the indices straight so the algebra matches the “next odd number” idea? A clear geometric interpretation would also help me see it.
Any help appreciated!
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3 Responses
Think of square numbers as a stack of “onion layers” made of little unit squares: start with a 1×1 core (that’s 1), then wrap it with an L-shaped layer of 3, then another L of 5, and so on. Why is each new layer 2n+1? If you have an n×n square and grow it to (n+1)×(n+1), you add one row of length n+1 and one column of length n; that’s (n+1)+n = 2n+1 new squares, with the corner counted exactly once because we deliberately take one side as n, the other as n+1 (the corner is shy and doesn’t like double invitations). So after n layers, you’ve added 1+3+5+…+(2n−1) squares, which exactly makes an n×n square: the sum is n^2. For the indices: the k-th odd number is 2k−1 if you start k at 1 (since 1=2·1−1, 3=2·2−1, …). The “next odd after 2n−1” is just 2 more, namely 2n+1, which is also 2(n+1)−1-so it’s the very next term, not two steps ahead. Algebraically that’s why S_{n+1} = S_n + (2n+1) = n^2 + 2n + 1 = (n+1)^2. It’s like leveling up a pixel-art square: each new ring has an odd count, and all the rings together paint a perfect square.
I love this pattern because it feels like building a square out of Lego: each new “layer” you wrap around the square is an odd-length strip. Start with an n×n square; to grow it to (n+1)×(n+1), you add an L-shaped border made of a new column of n squares and a new row of n+1 squares that meet at exactly one corner, so the total new squares are n + (n+1) = 2n + 1-an odd number. That’s the geometric reason the jumps between consecutive squares are odd: (n+1)² − n² = 2n + 1. Now the indexing: if you count odd numbers starting at k = 1, the k-th odd is 2k − 1 (so 1, 3, 5, …); the “next odd” after 2n − 1 is just (2n − 1) + 2 = 2n + 1, which is exactly the (n+1)-th odd, so the induction step lines up perfectly. Another tidy way to see it is to notice k² − (k−1)² = 2k − 1; summing these differences from k = 1 to n telescopes to n², which is the same as adding the first n odds. Simple worked example: start at a 1×1 square (=1), add an L of 3 to get 2×2 (=4), add an L of 5 to get 3×3 (=9), add an L of 7 to get 4×4 (=16), so 1 + 3 + 5 + 7 = 16. So both the picture and the algebra are saying the same thing: the first n odd numbers exactly “wrap” an n×n square, giving n².
I always remind myself: the k-th odd is 2k−1, so the “next odd” after 2n−1 is 2(n+1)−1 = 2n+1; adding it to 1+3+…+(2n−1)=n^2 gives n^2+(2n+1)=(n+1)^2, which the picture shows by growing an n×n square with an L of 2n+1 unit squares (n along the bottom, n along the right, plus the corner once).
Example: 3×3=9, add 2·3+1=7 to make 4×4=16, so 1+3+5+7=16.