When proving by induction, your first step is to show that a given claim is true for some minimal case (often called the base case). The given statement is a claim about the sum of numbers of the form 2n + 1 where n is a positive natural number. The least positive natural number is n = 1, so that seems like a reasonable place to start.
Now, we show the claim holds for n = 1:
3 + 5 + 7 + … + (2n + 1) = n (n + 2)
is a statement about an arbitrary positive integer n; the case of n = 1 reduces to
3 = 1 (1 + 2) = 1 • 3 = 3
which is obviously true.
Next, we assume that the statement is true for some arbitrary positive integer n = k, so that
3 + 5 + 7 + … + (2k + 1) = k (k + 2)
This is called the induction hypothesis.
We then use this assumption to show that the claim is also true for the next positive integer, n = k + 1. This is the claim that
3 + 5 + 7 + … + (2k + 1) + (2 (k + 1) + 1) = (k + 1) (k + 1 + 2)
or
3 + 5 + 7 + … + (2k + 1) + (2k + 3) = (k + 1) (k + 3)
By the induction hypothesis, we have
3 + 5 + 7 + … + (2k + 1) + (2k + 3)
= k (k + 2) + (2k + 3)
= k² + 2k + 2k + 3
= k² + 4k + 3
= (k + 1) (k + 3)
and this is exactly what we wanted to show, that if P(k) is true, then P(k + 1) is also true.
This means we have
P(1) = true ⇒ P(2) = true ⇒ P(3) = true ⇒ …
and so on, and the claim is true for all positive integers n.