Solution:

Let's list the numbers formed by repeating a single digit and calculate their digit sums:

1. Number: 11

   Digit sum: $1+1=2$

2. Number: 22

   Digit sum: $2+2=4$

3. Number: 33

   Digit sum: $3+3=6$

4. Number: 44

   Digit sum: $4+4=8$

5. Number: 55

   Digit sum: $5+5=10$

6. Number: 66

   Digit sum: $6+6=12$

7. Number: 77

   Digit sum: $7+7=14$

8. Number: 88

   Digit sum: $8+8=16$

9. Number: 99

   Digit sum: $9+9=18$

Pattern Observation:

The digit sums are 2, 4, 6, 8, 10, 12, 14, 16, 18.

This sequence consists of even numbers. More specifically, if the repeated digit is $d$, the number is $10d+d=11d$. The digit sum is $d+d=2d$.

So, the digit sum is always twice the repeated digit. This means the digit sums are consecutive even numbers, starting from $2\times 1=2$ up to $2\times 9=18$.

Final Answer: The digit sums are 2, 4, 6, 8, 10, 12, 14, 16, 18. The pattern is that the digit sum is always twice the repeated digit, resulting in a sequence of consecutive even numbers.