Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 1 Exercise 1.2

Solution: The $n$-th proposition is

$$

P_n: 3+11+\cdots+(8n-5)=4n^2-n.

$$ Then $P_1$ asserts $3=4\cdot 1^2-1$ 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:1-1-2-1}

3+11+\cdots+(8n-5)=4n^2-n.

\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:1-1-2-1} by $\big(8(n+1)-5\big)$ and obtain

\begin{align*}

&\ 3+11+\cdots+(8n-5)+\big(8(n+1)-5\big)\\

=&\ 4n^2-n+\big(8(n+1)-5\big)\\

=&\ 4n^2-n+8n+8-5\\

=&\ 4n^2+8n+4-(n+1)\\

=&\ 4(n^2+2n+1)-(n+1)\\

=&\ 4(n+1)^2-(n+1).

\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 $n$.