Writing a proper inductive argument The algebra is simple for some reason I am having a hard time setting up the argument. Here is my attempt
Suppose we have:
$\sum_{k=1}^n (2k-1)^2 = \frac{n(4n^2-1)}{3}$ for all $n \in \mathbb{N}$
Let see if this holds true for $n = 1$
LHS: $= 1$ 
RHS $= 1$
Thus this formula holds for $n=1$
Lets see if this formula holds true for $n+1$
$\sum_{k=1}^{n+1} (2k-1)^2 = ((2(n+1)-1)^2) + \sum_{k=1}^n (2k-1)^2$ 
$ = (2n+2-1)^2 + \frac{n(4n^2-1)}{3}$
Then after all the calculations I show that:
$(2n+2-1)^2 + \frac{n(4n^2-1)}{3}$ is equal to $\frac{n(4n^2-1)}{3}$ with $n+1$ in place of $n$ meaning:
$(2n+2-1)^2 + \frac{n(4n^2-1)}{3} = \frac{(n+1)(4(n+1)^2-1)}{3}$

Given the statement in this question, I will assume that by the natural numbers, you mean the set of positive integers.  If you instead meant the set of nonnegative integers, you need to prove that the statement holds when $n = 0$ as well, as shown by Antoine Mathys in the comments.
In writing a proof by mathematical induction about a property of the positive integers, you should:


*

*State the property $P(n)$ you wish to prove.

*Prove the base case $P(1)$ to establish that the property holds for $n = 1$. 

*State the induction hypothesis $P(m)$.

*Prove that $P(m) \implies P(m + 1)$ for each positive integer $m$.

*Conclude that since $P(1)$ holds and $P(m) \implies P(m + 1)$ for each positive integer $m$, then $P(n)$ holds for each positive integer $n$.


If you are instead proving a statement $P(n)$ for all nonnegative integers $n$, replace $1$ by $0$ and positive integer by nonnegative integer.  More generally, if you wish to prove a statement $P(n)$ holds for each integer $n \geq n_0$, replace $1$ by $n_0$ and show that $P(m) \implies P(m + 1)$ for each integer $m \geq n_0$.
Let $P(n)$ be the statement
$$\sum_{k = 1}^{n} (2k - 1)^2 = \frac{n(4n^2 - 1)}{3}$$
We wish to prove that this statement for each positive integer $n$.  
Proof.  Let $n = 1$.  Then 
$$\sum_{k = 1}^{1} (2k - 1)^2 = (2 \cdot 1 - 1)^2 = (2 - 1)^2 = 1^2 = 1 = \frac{1 \cdot 3}{3} = \frac{1(4 \cdot 1 - 1)}{3} =\frac{1(4 \cdot 1^2 - 1)}{3}$$
Thus, $P(1)$ holds.  
Since $P(1)$ holds, we may assume there is some positive integer $m$ such that 
$P(m)$ holds.  Then
$$\sum_{k = 1}^{m} (2k - 1)^2 = \frac{m(4m^2 - 1)}{3}$$
This is our induction hypothesis.
Let $n = m + 1$.  Then
\begin{align*}
\sum_{k = 1}^{m + 1} (2k - 1)^2 & = \sum_{k = 1}^{m} (2k - 1)^2 + \sum_{k = m + 1}^{m + 1} (2k - 1)^2\\
& = \sum_{k = 1}^{m} (2k - 1)^2 + [2(m + 1) - 1]^2\\
& = \sum_{k = 1}^{m} (2k - 1)^2 + [2m + 2 - 1]^2\\
& = \sum_{k = 1}^{m} (2k - 1)^2 + (2m + 1)^2\\
& = \sum_{k = 1}^{m} (2k - 1)^2 + 4m^2 + 4m + 1\\
& = \frac{m(4m^2 - 1)}{3} + 4m^2 + 4m + 1 && \text{by the induction hypothesis}\\
& = \frac{m(4m^2 - 1)}{3} + \frac{12m^2 + 12m + 3}{3}\\
& = \frac{4m^3 - m + 12m^2 + 12m + 3}{3}\\
& = \frac{4m^3 + 12m^2 + 11m + 3}{3}\\
& = \frac{(m + 1)(4m^2 + 8m + 3)}{3}\\
& = \frac{(m + 1)(4m^2 + 8m + 4 - 1)}{3}\\
& = \frac{(m + 1)[4(m^2 + 2m + 1) - 1]}{3}\\
& = \frac{(m + 1)[4(m + 1)^2 - 1]}{3}
\end{align*}
Thus, $P(m) \implies P(m + 1)$ for each positive integer $m$.
Since $P(1)$ holds and $P(m) \implies P(m + 1)$ for each positive integer $m$, then $P(n)$ holds for each positive integer $n$.$\blacksquare$
