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!
 A: You have proven that the base case works.
Induction Step: Assume that the claim holds true for $n=k$.
It remains to prove the claim true for $n=k+2$. Observe that:
$$ \begin{align*}
&1^2-3^2+5^2-7^2+\dots+(2k-3)^2-(2k-1)^2+(2(k+2)-3)^2-(2(k+2)-1)^2 \\
&= [1^2-3^2+5^2-7^2+\dots+(2k-3)^2-(2k-1)^2]+[(2k+1)^2-(2k+3)^2]  \\
&= [-2k^2]+[(2k+1)^2-(2k+3)^2] \qquad \text{by the induction hypothesis}\\
&= [-2k^2]+[((2k+1)+(2k+3))((2k+1)-(2k+3))] \qquad \text{difference of squares}\\
&= [-2k^2]+[(4k+4)(-2)] \\
&= -2(k^2)-2(4k+4) \\
&= -2(k^2+4k+4) \\
&= -2(k+2)^2 \\
\end{align*} $$
as desired. This completes the induction.
A: 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.
A: For $n=4$ you need to show  that $1^2-3^3+5^2-7^2 = -2\cdot 16$.
A: If $f(m)=1^2-3^2+5^2-7^2+\dots+(2m-3)^2-(2m-1)^2$
$m=2\implies f(2)=1^2-3^2=-8=-2.2^2$
$m=4\implies f(4)=1^2-3^2+5^2-7^2=1-9+25-49=-8-24=-2\cdot4^2$
