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

Question

$ 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 1

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)?

Answer 2

It seems like the process of induction is what you are having trouble with. For induction, you first prove that the statement holds true for n = 1. Then you assume it to be true for n = k. Then take n = k+1. Using your assumption, prove that the statement works for n=k+1. Once you have proven this, given that you have proven the statement for n=1, it must be true for k+1 (2) ,and so on for all the positive integers