Let,

\[~P\text{ }\left( n \right):\text{ }1\text{ }+\text{ }3\text{ }+\text{ }5\text{ }+\text{ }\ldots \text{ }+\text{ }\left( 2n\text{ }\text{ }1 \right)\text{ }=\text{ }{{n}^{2}}\]

Let us check P (n) is true for n = 1

\[P\text{ }\left( 1 \right)\text{ }=\text{ }1\text{ }={{1}^{2}}\]

\[1\text{ }=\text{ }1\]

P (n) is true for \[n\text{ }=\text{ }1\]

Now, Let’s check P (n) is true for \[n\text{ }=\text{ }k\]

\[P\text{ }\left( k \right)\text{ }=\text{ }1\text{ }+\text{ }3\text{ }+\text{ }5\text{ }+\text{ }\ldots \text{ }+\text{ }\left( 2k\text{ }\text{ }1 \right)\text{ }=\text{ }{{k}^{2}}~\ldots \text{ }\left( i \right)\]

To show

\[1\text{ }+\text{ }3\text{ }+\text{ }5\text{ }+\text{ }\ldots \text{ }+\text{ }\left( 2k\text{ }\text{ }1 \right)\text{ }+\text{ }2\left( k\text{ }+\text{ }1 \right)\text{ }\text{ }1\text{ }=\text{ }{{\left( k\text{ }+\text{ }1 \right)}^{2}}\]

Now,

\[1\text{ }+\text{ }3\text{ }+\text{ }5\text{ }+\text{ }\ldots \text{ }+\text{ }\left( 2k\text{ }\text{ }1 \right)\text{ }+\text{ }2\left( k\text{ }+\text{ }1 \right)\text{ }\text{ }1\]

\[=\text{ }{{k}^{2}}~+\text{ }\left( 2k\text{ }+\text{ }1 \right)\]

\[=\text{ }{{k}^{2}}~+\text{ }2k\text{ }+\text{ }1\]

\[=\text{ }{{\left( k\text{ }+\text{ }1 \right)}^{2}}\]

P (n) is true for \[n\text{ }=\text{ }k\text{ }+\text{ }1\]

Hence, P (n) is true for all n ∈ N.