Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 1 Exercise 1.3

Solution: The $n$-th proposition is

$$

P_n: 1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2.

$$ Then $P_1$ asserts $1^3=1^2$ 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:1-1-3-1}

1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2.

\end{equation} We would like to show $P_{n+1}$ is true based on $P_n$. To do that, we shall use the following identity from Example 1

\begin{equation}\label{eq:1-1-3-2}

1+2+\cdots+n=\frac{1}{2}n(n+1).

\end{equation} We add both sides of \eqref{eq:1-1-3-1} by $(n+1)^3$ and obtain

\begin{align*}

&\ 1^3+2^3+\cdots +n^3+(n+1)^3\\

=&\ (1+2+\cdots+n)^2+(n+1)^3\\

\text{use \eqref{eq:1-1-3-2}}=&\ \frac{1}{4}n^2(n+1)^2+(n+1)^3\\

=&\ \frac{1}{4}n^2(n+1)^2+\frac{1}{4}(n+1)^2(4n+4)\\

=&\ \frac{1}{4}(n+1)^2(n^2+4n+4)\\

=&\ \frac{1}{4}(n+1)^2(n+2)^2\\

=&\ \left(\frac{1}{2}(n+1)\big((n+1)+1\big)\right)^2\\

\text{use \eqref{eq:1-1-3-2}}=&\ \big(1+2+\cdots+n+(n+1)\big)^2.

\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 $n$.