An unbiased coin is tossed 12 times .The probability that at least 7 consecutive heads show up is $\frac{K}{256}$ then K =

The sequence of consecutive heads may starts with $1^{st}$ toss or $2^{nd}$ toss or $3^{rd}$ toss ---- or at $6^{th}$ toss. In any case ,if it starts with r th  throw , the first (r-2) throws may be head or tail but (r-1)st throw must be  tail, after which again a head or tail can show up:

$\therefore$ Probability $=\frac{1}{2^7}+\frac{1}{2}.\frac{1}{2^7}+\frac{1}{2}.\frac{1}{2^7}+------+\frac{1}{2}.\frac{1}{2^7}=\frac{1}{2^7}[1+\frac{5}{2}]=\frac{7}{2^8}$