Prove that the sum of consecutive odd numbers beginning at 1 (eg 1, 3, 5, ..) always adds up to a perfect square

For all integers n,
  (n + 1)² = n² + 2n + 1 = (n²) + (2n + 1)

For n = 1,
  (1 + 1)² = (1)² +2(1) + 1 = 1 + 3
For n = 2,
  (2 + 1)² = (2)² +2(2) + 1 = 4 + 5
For n = 3,
  (3 + 1)² = (3)² +2(3) + 1 = 9 + 7

For all integers n,
  (n + 1)² = n² + 2n + 1.
  → (n + 1)² - n² = 2n + 1

For n = 1,
  (2)² - (1)² = 2(1) +1 = 3
For n = 2,
  (3)² - (2)² = 2(2) +1 = 5
For n = 3,
  (4)² - (3)² = 2(3) +1 = 7

I forget what this method of proof is called or whether I really finished the proof, but I know I'm right. =]

Posted by Charley on 2005-05-14 09:37:07