- #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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