Check through option, condition $(n!{)}^2$ > $n^n$ is

true when n ≥ 3.

Check through option, the condition $3^n$ > $n^3$ is

true when n ≥ 4.

Putting n = 1 in $7^{2n}+2^{3n-3}{.3}^{n-1}$

=50, divisible by 25

$n(n^2-1)=(n-1)\left(n\right)(n+1)$

It is the product of three consecutive natural numbers. For any three consecutive natural numbers, one of them is divisible by 3 and at least one is divisible by 2. Hence, the product is always divisible by 6.

Check through option, the condition $2^n$ < n! is

true when n ≥ 4.

Let n = 1 then option (a) and (d) is eliminated.

Equality can't attained for any value of n so,

option (b) satisfied.

Check through option, the condition

${10}^{n-2}$ > 81n is satisfied if n ≥ 5.

Let n = 1, then option (a), (b) and (d)

eliminated. Only option (c) satisfied.

$x^{2n-1}+y^{2n-1}$ is always contain equal odd power.

So it is always divisible by x + y.

Check through option, the condition

$2^n$(n-1)!<$n^n$ is satisfied for n > 2.