Moment of inertia of a rod and ball system

I have a problem in which a rod of length $d$ has mass $m$ and a point mass of $2m$ is on the left end of it. I want to calculate the moments of inertia for several axes all perpendicular to the rod. I've already correctly calculated the moment of inertia directly for the axis through the rod's mid-point, finding it to be $\frac{7md^2}{12}$. But then it occurred to me that this would be more efficient if I found the moment of inertia through the center of mass and use the Parallel Axis Theorem repeatedly.

To do that I calculated the center of mass of the system and found it to be $d/6$. I then calculated the moment of inertia through the center of mass in two parts, one for the mass of the rod itself,

$$\int_{-d/6}^{5d/6}r^2dm = \lambda \frac{r^3}{3}\Bigg|_{-d/6}^{5d/6} = \frac{m}{3d}\left[\left(\frac{5d}{6}\right)^3-\left(\frac{-d}{6}\right)^3\right]$$

$$=\frac{md^2}{3}\left(\frac{126}{6^3}\right)=\frac{7md^2}{36}$$

But then I have to add the mass of the ball at the end which contributes $2m(d/6)^2$ to the moment of inertia and therefore the moment at the center is

$$I_{cm}=\frac{md^2}{4}$$

Now, however, when I use the parallel axis theorem to get the moment of inertia at the half-way point, I get

$$\frac{md^2}{4}+m(2d/6)^2 = \frac{13md^2}{36}$$

I clearly don't get the right answer.

When you apply the parallel axis theorem in the last line, you need to substitute $3m$ for mass in the second term of your equation, as $m + 2m$ is the total mass of your system. (Similarly, $\frac{1}{4}md^2$ is the total moment of inertia of the system about an axis through the rod's midpoint.)
After doing so, your equation evaluates to $\frac{1}{4}md^2 + \frac{1}{3}md^2 = \frac{7}{12}md^2$.