College Algebra, Chapter 9, 9.5, Section 9.5, Problem 6

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

By using mathematical induction,


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

So, $\displaystyle P(1) = \frac{4 (1 + 1)(2(1) + 1)}{6} = \frac{(2)(3)}{6} = \frac{6}{6} = 1$. Thus, we prove the first principle of the mathematical induction.

More over, assuming that $P(k)$ is true, then $\displaystyle 1^2 + 2^2 + 3^2 + ... k^2 = \frac{k (k + 1)(2k + 1)}{6}$

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


$
\begin{equation}
\begin{aligned}

1^2 + 2^2 + 3^2 + ... k^2 + (k + 1)^2 =& \frac{(k + 1) [(k + 1) + 1][2 (k + 1) + 1]}{6}
\\
\\
1^2 + 2^2 + 3^2 + ... k^2 + (k + 1)^2 =& \frac{(k + 1)(k + 2)(2k + 3)}{6}
\\
\\
1^2 + 2^2 + 3^2 + ... k^2 + (k + 1)^2 =& \frac{(k^2 + 3k + 2) (2k + 3)}{6}
\\
\\
1^2 + 2^2 + 3^2 + ... k^2 + (k + 1)^2 =& \frac{2k^3 + 6k^2 + 4k + 3k^2 + 9k + 6}{6}
\\
\\
1^2 + 2^2 + 3^2 + ... k^2 + (k + 1)^2 =& \frac{2k^3 + 9k^2 + 13k + 6}{6}
\\
\\
1^2 + 2^2 + 3^2 + ... k^2 + (k + 1)^2 =& \frac{1}{3} k^3 + \frac{3}{2} k^2 + \frac{13}{6} k + 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}

=& [1^2 + 2^2 + 3^2 + k^2] + [(k + 1)^2]
&& \text{Group the first $k$ terms}
\\
\\
=& \frac{k (k + 1)(2k + 1)}{6} + (k + 1)^2
&& \text{Induction hypothesis}
\\
\\
=& \frac{2k^3 + 3k^2 + k}{6} + k^2 + 2k + 1
&& \text{Expand}
\\
\\
=& \frac{1}{3} k^3 + \frac{1}{2} k^2 + \frac{1}{6} k + k^2 + 2k + 1
&&
\\
\\
=& \frac{1}{3} k^3 + \frac{3}{2} k^2 + \frac{13}{6} k + 1

\end{aligned}
\end{equation}
$


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