proof verification - Proving $0 + 2 + 4 + cdots + (2n-2) = n(n-1)$ for all n $in Z^+$

$0 + 2 + 4 + \cdots + (2n-2) = n(n-1)$ for all n $\in Z^+$

What I have:

Let n exist in Z^+. If n = 2, then L.H.S. = 2, and the R.H.S. = 2(2-1) = 2. So, L.H.S. = R.H.S. and this holds for n = 2. Assume this holds for some k existing in Z^+. That is 0 + 2 + 4 + ... + 2(k-2) = k(k-1)

What I am stuck on:

How do I show that this holds for 0 + 2 ... (2k-2) = k(k-1)?

Answer

You assume that for $n=k$, the statement holds:
$$0+2+\cdots+2(k-1) = k(k-1). \tag{a}$$
You then want to prove the statement for $n=k+1$, that is, we want to show
$$0+2+\cdots+2(k-1)+2k = (k+1)k.\tag{b}$$

Can you make the connection from (a) to (b)?