The term independent of x in the expression of $$\left(2x^2 + \frac{1}{x^2}\right)$$ is

we know that general term of expansion (a+b)^n is 

$$ T_{r+1}$$ = $$nC_{r}$${a}^{n-r}b^n       x$$\geq0$$

 here in expression  $$\left(2x^2 + \frac{1}{x^2}\right)$$  we have given   n=1    a=2x^2     b=1\x^2

$$ T_{r+1}$$ = $$1C_{r}$${2x^2}^{1-r}{1\x^2}^1

                   =   $$1C_{r}$${2}^{1-r}{x^2}^{1-r}{1\x^2}^1

                   =    $$1C_{r}$${2}^{1-r}{x}^{2-2r}{x}^-2r

                  =     $$1C_{r}$${2}^{1-r}{x}^-4r

so  from  that   n-r =0  and  r=4      so     n=$$5^{th}$$ term      answer