The probability that N throws contains (K-1) '6's somewhere in the first (N-1) throws, with a '6' on the N-th throw is $\displaystyle \binom{N-1}{K-1} \left(\frac56\right)^{(N-1)-(K-1)} \left(\frac16\right)^{K-1} \frac16$.
I'm not quite sure what you want to prove here. This sum has to be 1
since with probability 1 you will eventually reach K sixes. So the sum
of the probabilities of the number of throws needed for this is 1.