7. 14 A fair coin is tossed ten times, let $\frac{i}{j}$ be the probability that heads do not appear consecutively, where $i$ and $j$ are coprime, find $i+j$.

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[Solution] When tossing a coin ten times, the possibilities where heads do not appear consecutively are as follows:
10 times all tails has 1 possibility;
9 times tails and 1 time heads has $C_{10}^{1}$ possibilities;
8 times tails and 2 times heads, the non-consecutive possibilities can only be by inserting the two heads into the gaps between the 8 tails, so there are $C_{9}^{2}$ possibilities;
Similarly, 7 times tails and 3 times heads have $C_{8}^{3}$ possibilities;
6 times tails and 4 times heads have $C_{7}^{4}$ possibilities;
5 times tails and 5 times heads have $C_{6}^{5}$ possibilities.

For no more than 4 times tails, it is impossible for heads to not appear consecutively.
Since there are $2^{10}=1024$ possible outcomes for ten coin tosses.
The required probability is
$$
\begin{aligned}
\frac{i}{j} & =\frac{1+C_{10}^{1}+C_{9}^{2}+C_{8}^{3}+C_{7}^{4}+C_{6}^{5}}{1024} \\
& =\frac{144}{1024} \\
& =\frac{9}{64},
\end{aligned}
$$

Thus, $\quad i+j=73$.