Question

given s(n) sum of first n natural no's then z(n)=2 s(n)+41 least value of n that gives z(n) is not prime?
a)z(n) always prime
b)6
c)7
d)41

Answer 1

(b)6
if I am right please mark the brainist

Answer 2

We need to recall the following formula.

• Sum of first $n$ natural numbers: $\frac{n(n+1)}{2}$
• Prime number: a number that is divisible only by itself and $1$.

Given:

Sum of first $n$ natural numbers: $s(n)$

$z(n)=2s(n)+41$

Simplify the equation.

$z(n)=2(\frac{n(n+1)}{2} )+41$

$z(n)=n(n+1)+41$

Option b) $6$

We get,

$z(n)=n(n+1)+41$

$z(6)=6(7)+41$

$z(6)=83$

$83$ is a prime number.

Hence, this option is incorrect.

Option c) $7$

We get,

$z(n)=n(n+1)+41$

$z(7)=7(8)+41$

$z(7)=127$

$127$ is a prime number.

Hence, this option is incorrect.

Option d) $41$

We get,

$z(n)=n(n+1)+41$

$z(41)=41(42)+41$

$z(41)=1763$

$1763$ is not a prime number.

Hence, this option is correct.

Option a) $z(n)$ always prime

From option d), we get that

$z(n)$ is not prime for $n=41$.

Hence, this option is incorrect.

Therefore, the correct option is d) $41$.