P(n) = $n^2$ + n. It is always odd (statement) but

square of any odd number is always odd and 

also, sum of odd number is always even. So

for no any 'n' for which this statement is true.

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.

Check through option, the condition

 ${\left(\frac{n+1}{2}\right)}^n$ ≥ n ! is true for n  ≥ 1.

Check through option, the condition

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

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

true when n ≥ 4.

n$(n^2-1)$ = (n - 1)(n)(n + 1)

It is product of three consecutive natural

numbers, so according to Langrange's theorem

it is divisible by 3 ! i.e., 6.

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

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

option (b) satisfied.

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

=50, divisible by 25

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

So it is always divisible by x + y.