Proving $1+5+9+\cdots+(4n+1) = (n+1)(2n+1)$ by induction (is there a typo?) 
Using mathematical induction, prove that $$1+5+9+\cdots+(4n+1) = (n+1)(2n+1).$$

I understand the steps to take in order to prove by induction. It is also to my understanding that step 1 would be to verify that $n=1$ is true. I don't believe that it is true here.
When $n = 1$, we have $4n + 1 = 5$ and $(n + 1)(2n + 1) = 6$.
Am I missing something or is this a typo on my practice final?
 A: It may help you to realize that
$$
1+5+9+\cdots+(4n+1) = (n+1)(2n+1)
$$
may actually be rewritten as
$$
\sum_{i=0}^n(4i+1)=(n+1)(2n+1).
$$
Thus, for $n\geq 0$, let $S(n)$ denote the statement
$$
S(n) : \sum_{i=0}^n(4i+1)=(n+1)(2n+1).
$$
Base case ($n=0$): $S(0)$ says that $4(0)+1=1=(0+1)(2(0)+1)$, and this is true.
Induction step: Fix some $k\geq 0$ and assume that $S(k)$ is true where
$$
S(k) : \sum_{i=0}^k(4i+1)=(k+1)(2k+1).
$$
To be shown is that $S(k+1)$ follows where
$$
S(k+1) : \sum_{i=0}^{k+1}(4i+1)=(k+2)(2k+3).
$$
Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\sum_{i=0}^{k+1}(4i+1) &= (4k+5)+\sum_{i=0}^k(4i+1)\tag{by defn. of $\Sigma$}\\[0.5em]
&= (4k+5)+(k+1)(2k+1)\tag{by $S(k)$, the ind. hyp.}\\[0.5em]
&= (4k+5)+(2k^2+3k+1)\tag{expand}\\[0.5em]
&= 2k^2+7k+6\tag{simplify}\\[0.5em]
&= (k+2)(2k+3),\tag{factor}
\end{align}
we end up at the right-hand side of $S(k+1)$, completing the inductive step. 
Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 0$. $\blacksquare$
A: No, it's fine. For $n = 1$, the last term is $4n + 1 = 5$, so the LHS is just $1 + 5 = 6$, which matches the RHS. In fact, you could even start your base case at $n = 0$.
