Blog

Solution: The $n$-th proposition is
$$
P_n: 1^2+2^2+\cdots+n^2=\frac{1}{6}n(n+1)(2n+1).
$$ Then $P_1$ asserts $1^2=\dfrac{1}{6}\cdot 1(1+1)(2\cdot 1+1)$ which is clearly true and we have the induction basis.

Now we assume $P_n$ is true, that is the equation
\begin{equation}\label{eq:1-1-1-1}
1^2+2^2+\cdots+n^2=\frac{1}{6}n(n+1)(2n+1).
\end{equation} We would like to show $P_{n+1}$ is true based on $P_n$. To do that, we add both sides of \eqref{eq:1-1-1-1} by $(n+1)^2$ and obtain
\begin{align*}
&\ 1^2+2^2+\cdots +n^2+(n+1)^2\\
=&\ \frac{1}{6}n(n+1)(2n+1)+(n+1)^2\\
=&\ \frac{1}{6}n(n+1)(2n+1)+\frac{1}{6}(n+1)(6n+6)\\
=&\ \frac{1}{6}(n+1)\big(n(2n+1)+6n+6\big)\\
=&\ \frac{1}{6}(n+1)(2n^2+n+6n+6)\\
=&\ \frac{1}{6}(n+1)(2n^2+7n+6)\\
=&\ \frac{1}{6}(n+1)(n+2)(2n+3)\\
=&\ \frac{1}{6}(n+1)\big((n+1)+1\big)\big(2(n+1)+1\big).
\end{align*} Therefore $P_{n+1}$ is true if $P_n$ is true. By the principle of mathematical induction, we make the conclusion that $P_n$ is true for all positive integers $n$.