TITLE: Prove that for $n \in \mathbb{N}, \sum\limits_{k=1}^{n} (2k+1) = n^{2} + 2n $
QUESTION [5 upvotes]: I'm learning the basics of proof by induction and wanted to see if I took the right steps with the following proof:
Theorem: for $n \in \mathbb{N},  \sum\limits_{k=1}^{n} (2k+1)  =  n^{2} + 2n $
Base Case: 
let $$ n = 1 $$ Therefore $$2*1+1 = 1^{2}+2*1 $$ Which proves base case is true.
Inductive Hypothesis:
Assume $$\sum_{k=1}^{n} (2k+1)  =  n^{2} + 2n $$
Then $$\sum_{k=1}^{n+1} (2k+1)  =  (n+1)^{2} + 2(n+1) $$
$$\iff (2(n+1) +1)+ \sum_{k=1}^{n} (2k+1)  =  (n+1)^{2} + 2(n+1) $$
Using inductive hypothesis on summation term:
$$\iff(2(n+1) +1)+ n^{2} + 2n  =  (n+1)^{2} + 2(n+1) $$
$$\iff 2(n+1) = 2(n+1)  $$
Hence for $n \in \mathbb{N},  \sum\limits_{k=1}^{n} (2k+1)  =  n^{2} + 2n $ Q.E.D.
Does this prove the theorem? Or was my use of the inductive hypothesis circular logic?

REPLY [4 votes]: Your proof is fine, but for me, it didn't look like the clearest presentation.   For instance,  I think it'd be better to drop the biconditionals, and for your inductive step just write
$$\begin{eqnarray}
\sum_{k=1}^{n+1} (2k+1) &=&  n^2 + 2n + 2(n+1) + 1 \\
&=& n^2 + 2n + 1 + 2(n + 1)\\
&=& (n+1)^2 + 2(n+1)\\
\end{eqnarray}$$