College Algebra, Chapter 9, 9.5, Section 9.5, Problem 10

Prove that the formula $\displaystyle 1^3 + 3^3 + 5^3 + ... + (2n - 1)^3 = n^2(2n^2 - 1)$ is true for all natural numbers $n$.

By using mathematical induction,

Let $P(n)$ denote the statement $\displaystyle 1^3 + 3^3 + 5^3 + ... + (2n - 1)^3 = n^2(2n^2 - 1)$.

Then, we need to show that $P(1)$ is true. So,


$
\begin{equation}
\begin{aligned}

1^3 =& (1)^2 (2 (1)^2 - 1)
\\
\\
1 =& (2 -1)
\\
\\
1 =& 1

\end{aligned}
\end{equation}
$


Thus, we prove the first principle of the mathematical induction. More over, assuming that $P(k)$ is true, then

$1^3 + 3^3 + 5^3 + ... + (2k - 1)^3 = k^2 (2k^2 - 1)$

Now, by showing $P(k + 1)$, we have


$
\begin{equation}
\begin{aligned}

1^3 + 3^3 + 5^3 + ... (2k - 1)^3 + [2(k + 1) - 1]^3 =& (k + 1)^2 [2 (k + 1)^2 - 1]
\\
\\
1^3 + 3^3 + 5^3 + ... (2k - 1)^3 + [2(k + 1) - 1]^3 =& (k + 1)^2 [2k^2 + 4k + 1]

\end{aligned}
\end{equation}
$


We start with the left side and use the induction hypothesis to obtain the right side of the equation:


$
\begin{equation}
\begin{aligned}

=& \left[ 1^3 + 3^3 + 5^3 + ... + (2k - 1)^3 \right] + \left[ [2(k + 1) - 1]^3 \right]
&& \text{Group the first $k$ terms}
\\
\\
=& k^2 (2k^2 - 1) + [2(k + 1)-1]^3
&& \text{Induction hypothesis}
\\
\\
=& k^2 (2k^2 - 1) + [2k + 1]^3
&& \text{Expand}
\\
\\
=& k^2 (2k^2 - 1) + 8k^3 + 12k^2 + 6k + 1
&& \text{Combine like terms}
\\
\\
=& 2k^4 - k^2 + 8k^3 + 12k^2 + 6k + 1
&&
\\
\\
=& 2k^4 + 8k^3 + 11k^2 + 6k + 1
&& \text{Factor by using synthetic division}
\\
\\
=& (k + 1)^2 (2k^2 + 4k + 1)
&&
\end{aligned}
\end{equation}
$


Therefore, $P(k+1)$ follows from $P(k)$, and this completes the induction step.