Pendulum with Dirac Comb excitation

$l\ddot{\theta}+b\dot{\theta}+g\theta =G,\sum _{-\mathrm{\infty}}^{\mathrm{\infty}}\delta (t-nT)$

where l, b, g, G are constants and $T=\frac{2\pi}{\omega}$.

Show that the resulting motion is given by

$\theta \left(t\right)=\frac{G}{Tl{\omega}^{2}}+\frac{2G\mathrm{cos}(\omega t-\frac{\pi}{2})}{Tb\omega}+$ [terms with frequencies $\ge 2\omega$]

and explain why the higher frequency terms are supressed.

My first take was to rearrange to

$\ddot{\theta}+\frac{b}{l}\dot{\theta}+{\omega}^{2}\theta =\frac{G}{l},\sum _{-\mathrm{\infty}}^{\mathrm{\infty}}\delta (t-nT)$

where $\omega =\sqrt{\frac{g}{l}}$

Then, taking the Laplace transform of both sides I got

$\mathrm{\Theta}\left(s\right)=\frac{G}{l,\sqrt{{\omega}^{2}-{\left(\frac{b}{2l}\right)}^{2}}},\frac{\sqrt{{\omega}^{2}-{\left(\frac{b}{2l}\right)}^{2}}}{{(s+\frac{b}{2l})}^{2}-({\left(\frac{b}{2l}\right)}^{2}-{\omega}^{2})},\frac{1}{1-{e}^{-sT}}$

which, as far as I'm concerned transforms to

$\theta \left(t\right)=\frac{G}{l,\sqrt{{\omega}^{2}-{\left(\frac{b}{2l}\right)}^{2}}}\sum _{n=0}^{\mathrm{\infty}},H(t-nT),{e}^{-\frac{b}{2l}(t-nT)}\mathrm{sin}\left(\sqrt{{\omega}^{2}-{\left(\frac{b}{2l}\right)}^{2}}(t-nT)\right)$

And, assuming that this is a correct form of the solution, I can't see how that is equivalent with the function given in the question. I reckon it has something to do with using Fourier series/transform instead? If so, I'm not sure how to do that. Or, is there a way to convert my solution into the given one?

I've been struggling with this for a good few days now, so any help would be much appreciated.