$\begin{align} & \text{Use the method of mathematical induction to establish} \\ & 1+3+5+\cdot \cdot \cdot +(2n-1)={{n}^{2}} \\ & n\text{ is assumed to be positive integer} \\\end{align}$
$\begin{align} & \text{Proving by induction} \\ & 1+3+5+---+{{(2n-1)}^{2}}={{n}^{2}} \\ & \text{When }n=1\text{ L}\text{.H}\text{.S }=1,\text{ }R.H.S.={{1}^{2}}=1 \\ & \text{When }n=2\text{ }L.H.S;\text{ }1+3=4;\text{ }R.H.S={{2}^{2}}=4 \\ & \text{The formula is valid for }n \\ & \text{Let assume }n=k \\ & 1+3+5+---+{{(2k-1)}^{2}}={{k}^{2}}----(i) \\ & \text{For when }n=k+1\text{ term} \\ & 1+3+5+---+{{(2k-1)}^{2}}+{{[2(k+1)-1]}^{2}} \\ & \underbrace{1+3+5+----{{(2k-1)}^{2}}}_{k}+(2k+1)---(ii) \\ & \text{Substitute }{{k}^{2}}\text{ for }1+3+5+---+(2k-1) \\ & {{k}^{2}}+(2k+1)={{k}^{2}}+2k+1={{(k+1)}^{2}} \\ & \text{So the formula is also valid for (}k+1)\text{ terms} \\ & \text{The formula is valid for all positive integers} \\\end{align}$