Solution:
Part a
If $n=1$, the sum is $1$.
If $n=2$, the sum is $1+3=4$.
If $n=3$, the sum is $1+3+5=9$.
If $n=4$, the sum is $1+3+5+7=16$.
Note that $1=1^2$, $4=2^2$, $9=3^2$, and $16=4^2$. It is natural to guess the answer should be $n^2$.
Part b
$$
1+3+\cdots+(2n-1)=n^2
$$ is true for all positive integers $n$.
The $n$-th proposition is
$$
P_n: 1+3+\cdots+(2n-1)=n^2.
$$ Then $P_1$ asserts $1=1^2$ 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:1-4-1}
1+3+\cdots+(2n-1)=n^2.
\end{equation} We would like to show $P_{n+1}$ is true based on $P_n$. We add both sides of \eqref{eq:1-4-1} by $2(n+1)-1$ and obtain
\begin{align*}
&\ 1+3+\cdots+(2n-1)+\big(2(n+1)-1\big)\\
=&\ n^2+\big(2(n+1)-1\big)=n^2+(2n+2-1)\\
=&\ n^2+2n+1=(n+1)^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 $n$.
