What is the first value of n for which `n^2+n+41` is not a prime? 1 (b) 10 (c) 20 (d) 40

The smallest value of natural number n, for which 2n + 1 is not a prime number, is 3 (b) 4 (c) 5 (d) None of these

Let a_(n)=(10^(n))/(n!) for n=1,2,3, Then the greatest value of n for which a_(n) is the greatest is: (A) 11 (B) 20 (C) 10 (D) 8

For what value fo n are 2^(n)-1 and 2^(n)+1 prime?

Using mathematical induction prove that n^(2)-n+41 is prime.

The sum of first n odd natural numbers is 2n-1 (b) 2n+1 (c) n^2 (d) n^2-1

The sum of first n odd natural numbers is 2n-1 (b) 2n+1 (c) n^2 (d) n^2-1

If the mean of first n natural numbers is (5n)/9 , then n= (a) 5 (b) 4 (c) 9 (d) 10

If the mean of first n natural numbers is (5n)/9 , then n= (a) 5 (b) 4 (c) 9 (d) 10

If n is an integer between 0 and 21, then the minimum value of n!(21-n)! is attained for n= a. 1 b. 10 c. 12 d. 20