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.