Rank: ? (179)
Member #: 7065

Here's a challenge which most people here should get fairly easily.

Prove (mathematically) that 1 + 3 + 5 + ... + 2n - 1 = n^2, for any natural number n.

First correct proof wins a cookie.

Rank: ? (883)
Member #: 3

Answer

^^ What they said.


...wait, did I ruin it?

Rank: ? (171)
Member #: 11947

well then.. you bastard. Anyway a similar problem posed by Nicomachus in AD100 prooved that:

1^3 + 2^3 + 3^3 + ... n^3 = (1 + 2 + 3 + ... n)^2

Can you, by induction?

Rank: ? (171)
Member #: 11947

heres a hint

first prove that:

1^3 = 1
2^3 = 3 + 5
3^3 = 7 + 9 + 11
4^3 = 13 + 15 + 17 + 19
et cetera.... this is the left side of the equation...

for the right side, their is a similar proof that needs to be solve (and is fairly elementary, so will be described below)

1^2 = 1
2^2 = 1 + 3
3^2 = 1 + 3 + 5
4^2 = 1 + 3 + 5 + 7
5^2 = 1 + 3 + 5 + 7 + 9

or... 1 + 3 + ... + (2*n - 1) = n^2
so, for the first part, we know up to 5... and we shall say n is true, so what about n+1?

1 + 3 + ... + (2*n - 1) + (2(n+1) - 1= (n+1)^2
1 + 3 + ... + (2*n - 1) + (2*n + 1) = (n+1)^2
(n)^2 + 2*n + 1 = (n+1)^2

QED (well sorta... )

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"The universe seems ... to have been determined and ordered in accordance with number, by the forethought and the mind of the creator of all things; for the pattern was fixed, like a preliminary sketch, by the domination of number preexistent in the mind of the world-creating God.?
-Nicomachus of Gerasa, Arithmetic I, ca. AD100)