# Changing formula inside summation

I probably forgot something about summations but can someone explain to me how do you go from $$\sum_{n=-\infty}^\infty \delta(\frac{t}{2}-n)$$ to $$2\sum_{n=-\infty}^\infty \delta(t-2n)$$ Thanks

Using $$\delta(ax) = \frac{1}{|a|}\delta(x)$$ you get $$\delta(\frac{t}{2}-n) = \delta(\frac12(t-2n)) = \frac{1}{1/2}\delta(t-2n) = 2\delta(t-2n).$$