Base case: For [tex]n=1[/tex], the left side is 2 and the right is [tex]2\cdot1^2=2[/tex], so the base case holds.
Induction hypothesis: Assume the statement is true for [tex]n=k[/tex], that is
[tex]2+6+10+\cdots+4k-2=2k^2[/tex]
We want to show that this implies truth for [tex]n=k+1[/tex], that
[tex]2+6+10+\cdots+4k-2+4(k+1)-2=2(k+1)^2[/tex]
The first [tex]k[/tex] terms on the left reduce according to the assumption above, and we can simplify the [tex]k+1[/tex]-th term a bit:
[tex]\underbrace{2+6+10+\cdots+4k-2}_{2k^2}+4k+2[/tex]
[tex]2k^2+4k+2=2(k^2+2k+1)=2(k+1)^2[/tex]
so the statement is true for all [tex]n\in\mathbb N[/tex].
