If P (n) is the statement

$P(n): n^{2}-n+41$ is prime.

Now,

$P(1)=1^{2}-1+41=41$ (prime)

$P(2)=2^{2}-2+41=4-2+41=43$ (prime)

$P(3)=3^{2}-3+41=9-3+41=47$ (prime)

$P(41)=41^{2}-41+41=1681$ (not prime)

Thus, we can say that $P(1), P(2)$ and $P(3)$ are true, but $P(41)$ is not true.