Answer
See the explanation below.
Work Step by Step
We use mathematical induction as follows:
Statement ${{S}_{1}}$ is;
$1=\left( 2-1 \right)$
After simplifying on the right, we obtain
${{n}^{2}}=1$. This true statement shows that ${{S}_{1}}$ is true.
Suppose ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression,
${{S}_{k}}=1+3+5+....+\left( 2k-1 \right)={{k}^{2}}$
Adding $\left( 2\left( k+1 \right)-1 \right)$ on both sides as:
$\begin{align}
& 1+3+5+\ldots +\left( 2\left( k+1 \right)-1 \right)={{k}^{2}}+\left( 2\left( k+1 \right)-1 \right) \\
& 1+3+5+\ldots +\left( 2\left( k+1 \right)-1 \right)={{k}^{2}}+\left( 2k+2-1 \right) \\
& 1+3+5+\ldots +\left( 2\left( k+1 \right)-1 \right)={{k}^{2}}+2k+1 \\
& 1+3+5+\ldots +\left( 2\left( k+1 \right)-1 \right)={{\left( k+1 \right)}^{2}}
\end{align}$
Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}=1+3+5+\ldots +\left( 2n-1 \right)={{n}^{2}}$ holds true
