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.