For a $p$ dimensional random vector, the mean vector is of length $p$ and the covariance matrix is $p \times p$
Full covariance: If every parameter is free, then there are $p(p+1)/2$ parameters; one way of seeing this is that there's a covariance for each pair, of which there are $$\binom{p}{2} = \frac{p!}{(p-2)! \times 2} = \frac{p(p-1)}{2}$$
then add the diagonal entries, of which there are $p$:
$$ p + \binom{p}{2} = p + \frac{p(p-1)}{2} = \frac{2p + p^2 - p}{2}
= \frac{p^2 + p}{2} = \frac{p(p+1)}{2}$$
Diagonal: this means all of the off-diagonal entries are zero, so the only free parameters are on the diagonal; there are $p$ of those.
Spherical: this means all of the off diagonal entries are 0 and the diagonal entries are all equal, so there's $1$ free parameter.
If you add in the mean parameters to be estimated, you add $p$ to each of those totals and you get $p + p(p+1)/2$, $2p$, and $p+1$, respectively.