Answer
$\text{(a)}$
\begin{align}
1+3+5+...+(2n-1) = \sum_{k=1}^n (2k-1)
\end{align}
$\text{(b) See explanation.}$
Work Step by Step
$\text{(a)}$
\begin{align}
1+3+5+...+(2n-1) = \sum_{k=1}^n (2k-1)
\end{align}
$\text{(b) Let us use part (a) and prove it by using math. induction.}$
$\text{1) It should be true for n = 1:}$
\begin{align}
& \sum_{k=1}^n (2k-1) = n^2 \\
& 2 - 1 = 1^2 \Rrightarrow True
\end{align}
$\text{2) Let it be true for some number k = n:}$
\begin{align}
1+3+5+...+(2n-1) = n^2
\end{align}
$\text{2) Then it should be true for k = n +1 as well:}$
\begin{align}
&1+3+5+...+(2n-1) +(2(n+1)-1) = (n+1)^2 \\
&n^2 + 2n + 2 -1 = n^2+2n + 1 \\
&n^2+2n + 1 = n^2+2n + 1 \Rrightarrow True
\end{align}
$\text{Thus, it is true for any n that}$
\begin{align}
\sum_{k=1}^n (2k-1) = n^2
\end{align}
