Significance of $1 + 2 + 3 . . . +2n$ within inductive proof

Question

I am going through the third edition of Spivak's Calculus, and an element of an induction problem is leaving me with great confusion.

The problem in question is the second problem of Chapter 2, part (i), which asks one to find a formula for $$\sum_{i = 1}^n (2i-1)= 1+3+5\;.\;.\;.+(2n-1)$$

The solution to this problem is eventually $n^2$, I do not know why. The hint given asks what this expression has to do with $1+2+3\;.\;.\;.+2n$. My questions are these:

Why does this statement matter to the problem? Additionally, how can the expression be true if it contains odd numbers when it should only produce even ones?

Thank you in advance for any solutions.

Answer 1

The hint is implying you should start by considering $$1 + 2 + 3 + \cdots + 2n = \left[ 1 + 3 + 5 + \cdots + (2n-1) \right] + \underline{~~~~~~~~~~}$$

Where I assume you already know what the left side sums to

Answer 2

Inductive Proof

$$1+2+3+\cdots+(2n-1)=n^2. \tag 1$$ Obviously, $(1)$ holds for $n=1$,which is trivial.

Let's make the inductive hypothesis as $(1)$ holds for $n=k$, namely, $$1+2+3+\cdots+(2k-1)=k^2.$$

Then, when $n=k+1$, we obtain \begin{align*} 1+2+3+\cdots+[2(k+1)-1]&=(1+2+3+\cdots+2k-1)+[2(k+1)-1]\\&=k^2+2k+1\\&=(k+1)^2. \end{align*}

which shows that $(1)$ also holds for $n=k+1.$ By the induction principle, we are done.