This is equivalent to finding the powers of $k$ with coefficient $28$ in the expansion of $(k^2+k^4+k^6+k^8+...)^2(k+k^2+k^3+k^4+...)$.
But this is $\left(\frac{k^2}{1-k^2}\right)^2\cdot\frac{k}{1-k}\ =\ k^5\cdot\left(\frac{1}{1-k}\right)^3\cdot\left(\frac{1}{1+k}\right)^2\ =\ k^5\cdot(1+k)\cdot\left(\frac{1}{1-k^2}\right)^3$
$=k^5(1+k)(\tbinom{2}{2}+\tbinom{3}{2}k^2+\tbinom{4}{2}k^4+\tbinom{5}{2}k^6+...)$, the last part having general term $\tbinom{j}{2}k^{2j-4}$
from which it is easy to see that, since $\tbinom{8}{2}=28$, the last part contains the term $28k^{12}$ and the whole result $28k^{17}+28k^{18}$. So the answer is $\mathrm{(D)}$.