s=1,200+2,199+3,198+4,197+...+199,2+200,1

The given sum is of the form:

\[ S = 1 \times 200 + 2 \times 199 + 3 \times 198 + \ldots + 199 \times 2 + 200 \times 1. \]

Notice that this is a sequence where each term is of the form \( n \times (201 - n) \) for \( n \) ranging from 1 to 200. Therefore, the sum can be expressed as:

\[
S = \sum_{n=1}^{200} n \times (201 - n).
\]

Expanding each term:

\[
S = \sum_{n=1}^{200} (201n - n^2).
\]

This expression can be split into two sums:

\[
S = 201 \sum_{n=1}^{200} n - \sum_{n=1}^{200} n^2.
\]

Let's calculate each of these sums:

1. The sum of the first 200 natural numbers is given by:

\[
\sum_{n=1}^{200} n = \frac{200 \times 201}{2} = 20100.
\]

2. The sum of the squares of the first 200 natural numbers is given by:

\[
\sum_{n=1}^{200} n^2 = \frac{200 \times 201 \times 401}{6} = 2686700.
\]

Now, substitute these into the expression for \( S \):

\[
S = 201 \times 20100 - 2686700.
\]

Calculating this gives:

\[
S = 4040100 - 2686700 = 1353400.
\]

So, the value of the sum is:

\[
\boxed{1,353,400}.
\]

Would you like a detailed breakdown of any steps? Here are some related questions you could explore:

1. How do you derive the formulas for the sum of natural numbers and their squares?
2. What happens if you change the sequence range (e.g., summing from 1 to 100 instead of 200)?
3. How would you approach this sum if it included cubes instead of squares?
4. Can this sum be visualized geometrically?
5. How would this formula change if the sum were reversed (e.g., \( 200 \times 1 + 199 \times 2 + \ldots \))?

**Tip:** Recognizing patterns in sequences often simplifies complex problems, reducing them to well-known formulas or manageable expressions.

Explore how to calculate the sum of products of consecutive numbers from 1 to 200 using mathematical formulas. This problem involves the sum of natural numbers and squares, demonstrating key algebraic concepts. Suitable for high school students studying advanced arithmetic and algebra.

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