The given expression is:

\[
3^2 - 4^2 + 5^2 - 6^2 + \cdots + 49^2 - 50^2
\]

This is an alternating series of squared numbers where the odd numbers are squared and added, and the even numbers are squared and subtracted. Let's represent the pattern more generally:

\[
(3^2 - 4^2) + (5^2 - 6^2) + \cdots + (49^2 - 50^2)
\]

Using the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\), we can simplify each pair of terms:

\[
(n^2 - (n+1)^2) = (n - (n+1))(n + (n+1)) = (-1)(2n + 1) = -(2n + 1)
\]

Thus, each pair of terms can be rewritten as:

\[
3^2 - 4^2 = -(2(3) + 1) = -7
\]
\[
5^2 - 6^2 = -(2(5) + 1) = -11
\]
\[
\vdots
\]
\[
49^2 - 50^2 = -(2(49) + 1) = -99
\]

The expression now becomes the sum of negative odd numbers:

\[
-(7 + 11 + 15 + \cdots + 99)
\]

This is an arithmetic series with the first term \(7\), the common difference \(4\), and the last term \(99\).

To find the sum of this arithmetic series, we use the formula for the sum of an arithmetic series:

\[
S_n = \frac{n}{2} \times (a + l)
\]
where:
- \(a = 7\) (first term),
- \(l = 99\) (last term),
- \(n\) is the number of terms.

To find \(n\), we use the formula for the \(n\)-th term of an arithmetic sequence:

\[
a_n = a + (n-1) \times d
\]
Setting \(a_n = 99\), we solve for \(n\):

\[
99 = 7 + (n-1) \times 4
\]
\[
99 - 7 = (n-1) \times 4
\]
\[
92 = (n-1) \times 4
\]
\[
n-1 = \frac{92}{4} = 23
\]
\[
n = 24
\]

Now, we can calculate the sum:

\[
S_{24} = \frac{24}{2} \times (7 + 99) = 12 \times 106 = 1272
\]

Since the entire sum was negative:

\[
\boxed{-1272}
\]

Would you like further clarification or a more detailed breakdown?

Here are 5 related questions you might find interesting:
1. How do you determine the sum of an arithmetic series?
2. What is the difference of squares formula and how is it used?
3. How would the result change if the sequence started at \(1^2 - 2^2\)?
4. Can this pattern be generalized to other ranges of squares?
5. How can alternating series be visualized or represented graphically?

**Tip:** For arithmetic series, always identify the first term, common difference, and number of terms to simplify calculations!

3^2 - 4^2 + 5^2 - 6^2 + ... + 49^2 - 50^2

Discover how to calculate the sum of an alternating series of squares, starting with 3^2 - 4^2 and ending with 49^2 - 50^2. Learn to apply the difference of squares formula and sum arithmetic sequences step-by-step, suitable for Grades 9-11.

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