<s>[INST] Prove with Induction for $n\in \mathbb{N}$ and $n$ is even for $1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $ I want to prove by indection, for $n\in\mathbb N$ even: 
$$1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $$
what I did first is to check the numbers, so if $n$ is even
lets take $n=2$ so $(2\cdot 2-3)^2-(2\cdot 2-1)^2=-2\cdot 4$
lets take $n=4$ so $(2\cdot 4-3)^2-(2\cdot 4-1)^2\neq-2\cdot 16$

 I did something wrong?
Thanks!
 [/INST] If $f(m)=1^2-3^2+5^2-7^2+\dots+(2m-3)^2-(2m-1)^2$
$m=2\implies f(2)=1^2-3^2=-8=-2.2^2$
$m=4\implies f(4)=1^2-3^2+5^2-7^2=1-9+25-49=-8-24=-2\cdot4^2$
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