- #1

Venomily

- 15

- 0

[tex]1 = 1[/tex]

[tex]1 - 2^2 = -(1+2)[/tex]

[tex]1 - 2^2 + 3^2 = (1+2+3)[/tex]

[tex]1^2 - 2^2 + 3^2 - 4^2 = -(1+2+3+4)[/tex]

and so on.

I have to prove that this relationship is true for all natural numbers. This is what I did:

clearly it is true for 1, 2, 3 and 4.

assume true for n odd:

[tex]1^2 - 2^2 + 3^2 - 4^2 ... + n^2 = (1 + 2 + +3 + 4... + n)[/tex]

tidying things up a bit and inducting (n+1) and (n+2) we can obtain this pattern:

[tex](1^2 - 1) + (3^2 - 3)... + ((n-1)^2 - (n-1)) + ((n+1)^2 - (n+1)) = (2^2 + 2) + (4^2 + 4) +... + (n^2 + n) + ((n+2)^2 + (n+2))[/tex]

The [tex]((n+1)^2 - (n+1))[/tex] from the LHS cancels with the [tex](n^2 + n)[/tex] on the RHS if you play around with it, therefore the equality holds for every n+2 given any n >= 4. The same argument can be applied to the case in which n is even, QED.

[tex]1 - 2^2 = -(1+2)[/tex]

[tex]1 - 2^2 + 3^2 = (1+2+3)[/tex]

[tex]1^2 - 2^2 + 3^2 - 4^2 = -(1+2+3+4)[/tex]

and so on.

I have to prove that this relationship is true for all natural numbers. This is what I did:

clearly it is true for 1, 2, 3 and 4.

assume true for n odd:

[tex]1^2 - 2^2 + 3^2 - 4^2 ... + n^2 = (1 + 2 + +3 + 4... + n)[/tex]

tidying things up a bit and inducting (n+1) and (n+2) we can obtain this pattern:

[tex](1^2 - 1) + (3^2 - 3)... + ((n-1)^2 - (n-1)) + ((n+1)^2 - (n+1)) = (2^2 + 2) + (4^2 + 4) +... + (n^2 + n) + ((n+2)^2 + (n+2))[/tex]

The [tex]((n+1)^2 - (n+1))[/tex] from the LHS cancels with the [tex](n^2 + n)[/tex] on the RHS if you play around with it, therefore the equality holds for every n+2 given any n >= 4. The same argument can be applied to the case in which n is even, QED.

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