The question in the image translates to:

"Calculate the sum of all natural numbers between 1 and 200 that are divisible by 8."

To solve this, we can follow these steps:

1. Identify the sequence of numbers divisible by 8 within the range.

   • The smallest number between 1 and 200 that is divisible by 8 is 8 itself.
   • The largest number less than or equal to 200 that is divisible by 8 is 200.

   So, the sequence of numbers divisible by 8 between 1 and 200 is: $8, 16, 24, \ldots, 200$

2. Recognize this sequence as an arithmetic sequence.

   • The sequence has a common difference of 8.
   • The first term $a = 8$.
   • The last term $l = 200$.

3. Calculate the number of terms in this sequence.

   • The formula for the $n$-th term of an arithmetic sequence is: $a_n = a + (n - 1)d$
   • Set $a_n = 200$, $a = 8$, and $d = 8$: $200 = 8 + (n - 1) \times 8$ $200 - 8 = (n - 1) \times 8$ $192 = (n - 1) \times 8$ $n - 1 = 24$ $n = 25$

   So, there are 25 terms in this sequence.

4. Calculate the sum of the sequence.

   • The sum $S_n$ of an arithmetic sequence is given by: $S_n = \frac{n}{2} \times (a + l)$
   • Substituting $n = 25$, $a = 8$, and $l = 200$: $S_{25} = \frac{25}{2} \times (8 + 200)$ $S_{25} = \frac{25}{2} \times 208$ $S_{25} = 25 \times 104$ $S_{25} = 2600$

Answer:

The sum of all natural numbers between 1 and 200 that are divisible by 8 is 2600.

Would you like further explanation or have any questions?

Here are some related questions to extend your understanding:

1. How would you find the sum of all numbers divisible by a different number within a range?
2. What changes if the range were larger, such as 1 to 500?
3. How do arithmetic sequences help in finding sums quickly?
4. What is the sum of all even numbers between 1 and 200?
5. How can you verify that a sequence is arithmetic?

Tip: For any arithmetic sequence, knowing the first term, last term, and the common difference allows you to quickly find both the number of terms and the sum of the sequence.