According to ques:
\[P\text{ }\left( n \right)\text{ }=\text{ }{{n}^{2}}~+\text{ }n\] is even and P (r) is true, then \[{{r}^{2}}~+\text{ }r\] is even
Let \[{{r}^{2}}~+\text{ }r\text{ }=\text{ }2k\text{ }\ldots \text{ }\left( i \right)\]
Now, \[{{\left( r\text{ }+\text{ }1 \right)}^{2}}~+\text{ }\left( r\text{ }+\text{ }1 \right)~\]
Or,
\[{{r}^{2}}~+\text{ }1\text{ }+\text{ }2r\text{ }+\text{ }r\text{ }+\text{ }1\]
or,
\[({{r}^{2}}~+\text{ }r)\text{ }+\text{ }2r\text{ }+\text{ }2\]
\[2k\text{ }+\text{ }2r\text{ }+\text{ }2\] [from equation (i)]
\[2\left( k\text{ }+\text{ }r\text{ }+\text{ }1 \right)\]
\[2\mu \]
\[\therefore ~{{\left( r\text{ }+\text{ }1 \right)}^{2}}~+\text{ }\left( r\text{ }+\text{ }1 \right)\] is Even.
Hence, \[P\text{ }\left( r\text{ }+\text{ }1 \right)\text{ }is\text{ }true.\]