College Algebra, Chapter 9, 9.5, Section 9.5, Problem 8

Prove that the formula $\displaystyle 1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5
+ ... + n(n + 2) = \frac{n(n + 1)(2n + 7)}{6}$ is true for all natural
numbers $n$.

By using mathematical induction,

Let $P(n)$ denote the statement $\displaystyle 1 \cdot 3 + 2 \cdot 4 +
3 \cdot 5 + ... + n(n + 2) = \frac{n(n + 1)(2n + 7)}{6}$.

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


$
\begin{equation}
\begin{aligned}

1 \cdot 3 =& \frac{(1)(1+1)(2(1)+ 7)}{6}
\\
\\
3 =& \frac{(2)(9)}{6}
\\
\\
3 =& \frac{18}{6}
\\
\\
3 =& 3

\end{aligned}
\end{equation}
$


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

$\displaystyle 1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + ... k(k + 2) =
\frac{k(k + 1)(2k + 7)}{6}$

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


$
\begin{equation}
\begin{aligned}

1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + ... k (k + 2) + (k + 1)[(k + 1) + 2]
=&\frac{(k + 1)[(k + 1) + 1][2(k + 1) + 7]}{6}
\\
\\
1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + ... k (k + 2) + (k + 1)(k + 3)
=& \frac{(k + 1)(k + 2)(2k + 9)}{6}
\\
\\
1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + ... k (k + 2) + (k + 1)(k + 3)
=& \frac{(k^2 + 3k + 2)(2k + 9)}{6}
\\
\\
1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + ... k (k + 2) + (k + 1)(k + 3)
=& \frac{2k^3 + 9k^2 + 6k^2 + 27k + 4k + 18}{6}
\\
\\
1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + ... k (k + 2) + (k + 1)(k + 3)
=& \frac{2k^3 + 15k^2 + 31k + 18}{6}
\\
\\
1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + ... k (k + 2) + (k + 1)(k + 3)
=& \frac{1}{3} k^ 3 + \frac{5}{2} k^2 + \frac{31}{6} k + 3
\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}

=& [1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + k (k + 2)] + [(k + 1)(k + 3)]
&& \text{Group the first $k$ terms}
\\
\\
=& \frac{k(k + 1)(2k + 7)}{6} + (k + 1)(k + 3)
&& \text{Induction hypothesis}
\\
\\
=& \frac{2k^3 + 9k^2 + 7k}{6}
&& \text{Expand}
\\
\\
=& \frac{1}{3} k^3 + \frac{3}{2} k^2 + \frac{7}{6} k + k^2 + 4k + 3
&& \text{Simplify}
\\
\\
=& \frac{1}{3} k^3 + \frac{5}{2} k^2 + \frac{31}{6} k + 3

\end{aligned}
\end{equation}
$


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