Given:

P(n) = n^{2} – n + 41 is prime.

P(n) = n^{2} – n + 41

P (1) = 1 – 1 + 41

= 41

P (1) is Prime.

Similarly,

P(2) = 2^{2} – 2 + 41

= 4 – 2 + 41

= 43

P (2) is prime.

Similarly,

P (3) = 3^{2} – 3 + 41

= 9 – 3 + 41

= 47

P (3) is prime

Now,

P (41) = (41)^{2} – 41 + 41

= 1681

P (41) is not prime

Hence, P (1), P(2), P (3) are true but P (41) is not true.

Answered by Aaryan | 1 year agoGiven an example of a statement P (n) such that it is true for all n ϵ N.

a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]

7^{2n} + 2^{3n – 3}. 3n – 1 is divisible by 25 for all n ϵ N

n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.

(ab)^{ n} = a^{n} b^{n} for all n ϵ N