can you check my proof by induction?

Venomily

[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.

Mentallic

If you're going to assume true for n, then on the very next line you have +n² as the last term on the LHS and a positive sum on the RHS, then you should specify "assume true for n odd" or something along those lines.

Also, it's unnecessary to take it in two cases, just set up your assumption as,

Assume true for n=k:

[tex]1^2 - 2^2 + 3^2 - ... + (-1)^{k-1}k^2=(-1)^{k-1}(1+2+3+...+k)[/tex]

Now prove it is true for n=k+1.

Venomily

@bold, how would you go about proving it then? rearranging the expression would be tedious with a multiplier (is there a better way?). I thought it would be easier to take the two cases since if you prove one case you imply the other.

EDIT: nvm, I found a more refined way of doing it using the (-1)^k-1, thanks.

Mentallic

If you use the well known summation property [tex]1+2+3+...+n=\frac{n(n+1)}{2}[/tex] it becomes very easy to solve.

Venomily

thanks, I forgot about using known results :p Yes, it becomes very easy now.