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 Odd Sum (Posted on 2002-08-08)
Prove that the sum of consecutive odd numbers beginning at 1 (eg 1, 3, 5, ..) always adds up to a perfect square

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 Arithmetic series | Comment 2 of 17 |
We can present the series of consecutive odd numbers as the following arithmetic series:
A1, A2, A3, A4, ... , An. (Ai belongs to N)
We can define Ai as Ai = 1 + Ri, where Ri = Ri-1 + 2 (i-1 is a sufix to R in the last equation).

Obviously, since the problem states that the series begins with 1, and we defined Ai to belong to N, it means that R1 has to be 0 in order to have A1 equal 1.

Also, if we observe the series carefully, we notice that the number of numbers in the series at any point equals:
n = (Rn/2) + 1
From this, we have that Rn = 2(n - 1).

(To prove this point we can show the following series representing n:
1 + (2/2) + (2/2) + ... + (2/2) where its sum is
S = n = 1 + 2(n-1)/2 (n-1 is the number of (2/2)s in the series of n).
The series of Ri can be shown as:
R1, R1 + 2, R2 + 2, ... , Rn-1 + 2 where R1 = 0.
From this we have that Rn = 2(n-1). If we put this in the above sum, we have proved the point)

The sum of the arithmetic series is:
S = (n/2)(A1 + An) = (n/2)(A1 + 1 + Rn) = (n/2)(A1 + 1 + 2n - 2) = (n/2)(A1 + 2n - 1).
Since we defined A1 as equalling 1 at the beginning of proof, the above equation comes to the following:
S = (n/2) x 2n = n²
 Posted by lucky on 2002-08-08 09:57:06

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