If P (n) is the statement

$P(n): n^{2}+n$ is even.

Also,

$P(r)$ is true.

Thus, $r^{2}+r$ is even.

To prove: $P(r+1)$ is true.

Now,

$P(r+1)=(r+1)^{2}+r+1$

$=r^{2}+1+2 r+r+1$

$=r^{2}+3 r+2$

$=r^{2}+r+2 r+2$

$=P(r)+2(r+1)$

$P(r)$ is even.

Also, $2(r+1)$ is even, $a s$ it is a multiple of 2 .

Therefore, $P(r+1)$ is even and true.