Let S(q) = sum for k=0 to infinity of q^k.

If q<1, this "geometric series" converges and is equal to 1/(1-q).

(e.g. S(1/2) = 1 + 1/2 + (1/2)^2 + (1/2)^3 + ... = 1 + 1/2 + 1/4 + 1/8 + ... = 1/(1-1/2) = 2)

The derivative of S(q) can be seen to equal

S'(q) = 1/(1-q)^2

It's also true that

S'(q) = sum for k=0 to infinity of k*q^(k-1)

So, sum for k=0 to infinity of k*q^(k-1) = 1/(1-q)^2

In our case, q=19/20.