Answer
$v = \sqrt{\frac{k~\Delta l~(l+\Delta l)}{m}}$
Work Step by Step
We can find the tension in the rubber band:
$\tau = k~\Delta l$
We can find the wave speed:
$v = \sqrt{\frac{\tau}{\mu}}$
$v = \sqrt{\frac{k~\Delta l}{m/(l+\Delta l)}}$
$v = \sqrt{\frac{k~\Delta l~(l+\Delta l)}{m}}$
