Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 1 Exercise 1.10
Solution: Recall from Exercise 1.4 that
\begin{equation}\label{eq:1-10-1}
1+3+\cdots+(2n-1)=n^2.
\end{equation} Note that
\begin{align*}
& (2n+1)+(2n+3)+\cdots+(4n-1)\\=&\ (1+3+5+\cdots+(4n-3)+(4n-1)) -(1+3+\cdots+(2n-1)).
\end{align*} Using \eqref{eq:1-10-1}, we find that
$$
1+3+5+\cdots+(4n-3)+(4n-1)=(2n)^2=4n^2,
$$ and
$$
1+3+\cdots+(2n-1)=n^2.
$$ Therefore
$$
(2n+1)+(2n+3)+\cdots+(4n-1)=4n^2-n^2=3n^2.
$$