The problem in the image asks to calculate the sum:

$111 + 112 + 113 + \dots + 212 + 222$

This is a sum of an arithmetic sequence where the first term $a = 111$ and the last term $l = 222$, with a common difference $d = 1$.

Step-by-step solution:

1. Determine the number of terms $n$: The number of terms in an arithmetic sequence can be found using the formula: $n = \frac{l - a}{d} + 1$ Substituting the known values: $n = \frac{222 - 111}{1} + 1 = 112$ So, there are 112 terms.

2. Use the sum formula for an arithmetic sequence: The sum $S_n$ of an arithmetic sequence is given by: $S_n = \frac{n}{2} \times (a + l)$ Substituting the values: $S_{112} = \frac{112}{2} \times (111 + 222) = 56 \times 333 = 18,648$

Final Answer:

$111 + 112 + 113 + \dots + 222 = 18,648$

Would you like more details or have any questions? Here are five related questions:

1. How do you identify the common difference in an arithmetic sequence?
2. What would the sum be if the sequence started from 100 instead of 111?
3. Can you derive the formula for the number of terms in any arithmetic progression?
4. What happens if the common difference changes to a value other than 1?
5. How does the sum formula change if only the first half of the sequence is taken?

Tip: Always verify the number of terms in an arithmetic sequence before applying the sum formula to avoid miscalculations.