I want to be independent!

Via the Binomial Theorem , the terms are of the form $\binom{8}{n} \cdot (2x)^{8-n} \cdot \left(\frac{k}{x}\right)^n.$

For the term to be independent of $x,$ we need

$x^{8-n}(\frac{1}{x})^n = x^0$

$x^{8-n}(x^{-1})^{n} =x^0$

$x^{8-n}x^{-n} = x^0$

$x^{8-n} = x^n$

$8-n =n$

$2n= 8$

$n=4$

Thus, we have a constant term of $\binom{8}{4} \times 2^4 \times k^4 = 700000$

$70\times 16 \times k^4 = 700000$

$1120k^4 = 700000$

$k^4 = \frac{700000}{1120} =625$

$k=\boxed{5}$

I tried expressing it in the form of $(1+x)^n$ .

$(\frac {k}{x})^8(1+\frac {2x^2}{k})$

For $x^8$ in the denominator to cancel out, we will need $x^8$ in the numerator. So we want the fifth term in the sequence.

$(\frac {k}{x})^8 \cdot \binom {8}{4} \cdot (\frac {2x^2}{k})^4 =70 \cdot 16 \cdot k^4=700000$

$k=5$