Relation between enthalpy and Cv

MY ATTEMPT

We already know that $$\Delta U=nC_V\,\Delta T$$ which is for constant volume

Now we know that $$\Delta H = \Delta U + p\,\Delta V$$

so $$\Delta H = nC_V\,\Delta T + 0$$ (as at constant volume $$\Delta V = 0$$)

$$\Delta H=nC_V\,\Delta T$$

But I have never such an equation anywhere.

For an ideal gas, your equation for $\Delta U$ applies not only to constant volume processes, but to all processes. It's just that, in experiments to measure Cv, it is very convenient to use a constant volume path because, under those circumstances, no work is done, in which case $\Delta U$ is just equal to the amount of heat added. So this gives you a direct measurement of Cv.
As far as $\Delta H$ is concerned, the correct equation is $\Delta H=\Delta U+\Delta (PV)$. But, from the ideal gas law, $\Delta (PV)=nR\Delta T$. So what does that give you for $\Delta H$?