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.