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?
On
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 $n\geq 0$. $\blacksquare$
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 $n = 0$.