Fifty Odd Numbers, One Square

Question

Compute 1 + 3 + 5 + cdots + 99, the sum of the first 50 odd numbers, in under ten seconds, then give two different proofs of the general pattern you used.

Accepted Answer

Answer: 50^2 = mathbf2500. The sum of the first n odd numbers is exactly n^2. Proof 1, telescoping. The k-th odd number is a difference of consecutive squares: k^2 - (k-1)^2 = 2k - 1. Summing over k = 1, ldots, n, the left side collapses to n^2 - 0, done. (The same collapsing-sum trick that evaluates sum frac1k(k+1); differences of a running quantity always telescope.) Proof 2, pairing / average. An arithmetic series sums to (average term) times (number of terms). The 50 odds from 1 to 99 have average frac1 + 992 = 50, so the sum is 50 times 50 = 2500. Geometrically: each odd number 2k - 1 is the L-shaped border that grows a (k-1)times(k-1) square into a k times k square, stack the borders and you've built the square. The transferable skill is the average-times-count reflex for any arithmetic ladder, which shows up constantly in sizing arithmetic: a scale-in that buys 100, 200, 300, ldots, 1000 shares across ten levels is 10 times 550 = 5500 shares total, one multiplication. Ladders of strikes, laddered orders, linear decay schedules, all of them sum at a glance once you see "average times count." The squares pattern itself is a useful audit: cumulative sums of a quantity that grow

Fifty Odd Numbers, One Square

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