Prove that $(2n+1)+(2n+3)+\dots +(4n-1) = 3n^2$ by induction Note: This is for self study, the book is Elementary analysis by Kenneth. A. Ross

How to prove the following by mathematical induction, I am stuck
Hint:
The $P_1$ step is correct. Your $P_{n+1}$ is not. Look at this hint:

$P_{n+1} = P_n - (2n+1) + (4n+1) + (4n+3) = 3n^2 +6n + 3$

This comes from:

$$\overbrace{2n+1 + 2n+3 + \ldots + 4n-1}^{P_n} + 4n+1 + 4n+3$$

$$2n+1 +  \underbrace{2n+3 + \ldots + 4n-1 + 4n+1 + 4n+3}_{P_{n+1}}$$
You've written $P_{n+1}$ incorrectly; what you want is
$$(2(n+1)+1) + (2(n+1)+3) + \cdots + (4(n+1)-1) \stackrel{?}{=} 3(n+1)^2.$$