I'm struggling to find the recurrence relation to evaluate the volume of a solid formed by a series of truncated cones one on top of the other. The image below illustrates the problem for 2 truncated cones, with the solid being formed by the complete revolution of the area around the $z$ axis. Each $i$ truncated cone is characterized by its height $H_i$ and the aperture angle $\alpha_i$, while the smallest radius is $R$ and $0\le\alpha<\frac{\pi}{2}$.
Since the volume of a truncated cone is given by: $$V = \frac{\pi H}{3}\left(r_1^2+r_1r_2+r_2^2\right)$$
For 1 truncated cone: $$V_1 = \frac{\pi H}{3}\left[R^2+R(R+H_1\tan\alpha_1)+(R+H_1\tan\alpha_1)^2\right]$$
For 2 truncated cones: $$V_2 = \frac{\pi}{3}\left\{H_1\left[R^2+R(R+H_1\tan\alpha_1) + (R+H_1\tan\alpha_1)^2\right] + H_2\left[(R+H_1\tan\alpha_1)^2 + (R+H_1\tan\alpha_1)(R+H_1\tan\alpha_1+H_2\tan\alpha_2)+(R+H_1\tan\alpha_1+H_2\tan\alpha_2)^2 \right]\right\}$$
And so on. Just by analyzing the expressions, my first thought was to propose the following formula for a series of $n$ truncated cones: $$V_n = \frac{\pi}{3}\sum_{i=1}^{n}{H_i\left[\left(R+H_i\tan\alpha_i+H_{i-1}\tan\alpha_{i-1}+H_{i-2}\tan\alpha_{i-2}+...\right)^2 +\left(R+H_i\tan\alpha_i+H_{i-1}\tan\alpha_{i-1}+H_{i-2}\tan\alpha_{i-2}+...\right)\left(R+H_{i-1}\tan\alpha_{i-1}+H_{i-2}\tan\alpha_{i-2}+...\right) + \left(R+H_{i-1}\tan\alpha_{i-1}+H_{i-2}\tan\alpha_{i-2}+...\right)^2 \right]} $$
But i'm not sure if this is a rigorous solution or the most elegant way to represent this volume. For example, if $i<1$, than both $H_i$ and $\alpha_i$ must be zero. Thanks in advance!