Solution

Self‑inductances

For a long solenoid, the total self‑inductance is $L = \mu_0 n^2 \pi r^2 l$
Therefore:

$L_1 = \mu_0 n_1^2 \pi r_1^2 l_1, \qquad L_2 = \mu_0 n_2^2 \pi r_2^2 l_2$

Mutual inductance

The mutual inductance M is defined from the flux that one coil sends through the other.
The magnetic field created by the outer solenoid (of length$l_2$) in its interior is uniform and is given by$B_2 = \mu_0 n_2 I_2$ This field passes through the cross‑section of the inner solenoid ($\pi r_1^2$) along the common length, which is $l_2$
The flux linked by the inner solenoid due to current $I_2$is:

$\Phi_{12} = (B_2)(\pi r_1^2)(n_1 l_2) = \mu_0 n_1 n_2 \pi r_1^2 l_2 I_2$

By definition, $M = \Phi_{12}/I_2$:

$M = \mu_0 n_1 n_2 \pi r_1^2 l_2$

Total inductance according to the direction of the currents

In a series connection the total inductance is:

$L = L_1 + L_2 \pm 2M$

where the + sign corresponds to currents in the same direction (fields reinforcing each other) and the - sign to currents in opposite directions (fields opposing each other).

Substituting the expressions:

$\boxed{L_{\pm} = \mu_0 \pi \Bigl( n_1^2 r_1^2 l_1 + n_2^2 r_2^2 l_2 \pm 2 n_1 n_2 r_1^2 l_2 \Bigr)}$