Solution:

On the $n$th straight path, the caterpillar travels $n$ units before hitting food and his weight is $n-1$. Then his speed is $2^{1-n}$. Then right before he turns around for the $n$th time, he has traveled a total time of $\sum_{i=1}^{n} \frac{i}{2^{1-i}}=\frac{1}{2} \sum_{i=1}^{n} i \cdot 2^{i}$. We want to know how many days the caterpillar moves before his weight is 10, so we want to take $n=10$ so that his last straight path was taken at weight 9. Hence we want to evaluate $S=\frac{1}{2} \sum_{i=1}^{10} i \cdot 2^{i}$. Note that $2S=\frac{1}{2} \sum_{i=2}^{11}(i-1) \cdot 2^{i}$, so $S=2S-S=\frac{1}{2}\left(11 \cdot 2^{11}-\sum_{i=1}^{10} 2^{i}\right)=\frac{1}{2}\left(10 \cdot 2^{11}-2^{11}+2\right)=9217$.