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!

Let's put $n=2m$ we can then write the sum as:
$$(1^2-3^2)+(5^2-7^2) +\dots +((4m-3)^2-(4m-1)^2) =$$$$(1-3)\cdot(1+3)+(5-7)\cdot(5+7)+ \dots +((4m-3)-(4m-1))\cdot((4m-3)+(4m-1))=$$$$-2\cdot4-2\cdot12-\dots - 2\cdot(8m-4)=$$$$-8(1+3+\dots(2m-1))$$
The sum (in brackets) is now a  linear one (well known to add to $m^2$ - proved by an easy induction, which I will leave to you). Since $-8m^2=-2n^2$ the desired result follows.
