The expression $1^2 - 2^2 + 3^2 - 4^2 + \dots + 135^2 - 136^2$ is an alternating series of squares. To solve this, let's first simplify the problem.

Step 1: Recognize the Pattern

Each pair of terms can be grouped together as: $n^2 - (n+1)^2$ This can be simplified using the difference of squares: $n^2 - (n+1)^2 = n^2 - (n^2 + 2n + 1) = -2n - 1$ So, each pair $n^2 - (n+1)^2$ simplifies to $-2n - 1$.

Step 2: Sum of the Series

The series can be rewritten as: $-(2 \times 1 + 1) - (2 \times 3 + 1) - \dots - (2 \times 135 + 1)$ Notice that each term in the sum is a negative odd number, starting from $-3$ to $-271$.

Step 3: Sum of Arithmetic Series

The sum of the series can be calculated as: $S = -\sum_{n=1}^{68} (2n + 1)$ Where $68$ is half of $136$ (because there are $68$ pairs of numbers).

This sum can be split as: $S = -\left( \sum_{n=1}^{68} (2n) + \sum_{n=1}^{68} 1 \right)$

• The first sum $\sum_{n=1}^{68} 2n$ is the sum of the first 68 even numbers: $\sum_{n=1}^{68} 2n = 2 \times \sum_{n=1}^{68} n = 2 \times \frac{68 \times 69}{2} = 68 \times 69 = 4692$
• The second sum is simply $68$ since you're adding $1$ for each term.

So, $S = -(4692 + 68) = -4760$

Final Answer

The sum of the series $1^2 - 2^2 + 3^2 - 4^2 + \dots + 135^2 - 136^2$ is $-4760$.

Would you like more details or have any questions? Here are some related questions to consider:

1. How does the difference of squares simplify algebraic expressions?
2. What are the properties of arithmetic series?
3. How can alternating series be used in calculus?
4. Why is it useful to recognize patterns in series like this one?
5. Can this method be generalized to other types of alternating series?

Tip: When dealing with alternating sums, grouping terms to identify patterns can significantly simplify the problem.