 solve 1^2-2^2+3^2-4^2+5^2-6^2 +...+135^2-136^2

The expression given is:

\[
1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + \dots + 135^2 - 136^2
\]

We can notice that the expression alternates between squares of consecutive integers with alternating signs. This allows us to group the terms in pairs:

\[
(1^2 - 2^2) + (3^2 - 4^2) + (5^2 - 6^2) + \dots + (135^2 - 136^2)
\]

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

\[
1^2 - 2^2 = (1-2)(1+2) = -1 \times 3 = -3
\]
\[
3^2 - 4^2 = (3-4)(3+4) = -1 \times 7 = -7
\]
\[
5^2 - 6^2 = (5-6)(5+6) = -1 \times 11 = -11
\]

In general, each term in the series becomes:

\[
(2n-1)^2 - (2n)^2 = -1 \times [(2n-1) + (2n)] = -4n
\]

So, the entire sum becomes:

\[
\sum_{n=1}^{68} -4n
\]

where \( n \) goes from 1 to 68 (since \( 135 = 2 \times 68 - 1 \) and \( 136 = 2 \times 68 \)).

The sum of the first \( N \) natural numbers is given by the formula:

\[
\sum_{n=1}^{N} n = \frac{N(N+1)}{2}
\]

Thus:

\[
\sum_{n=1}^{68} n = \frac{68 \times 69}{2} = 2346
\]

Multiplying by \(-4\), the sum is:

\[
-4 \times 2346 = -9384
\]

So, the value of the expression is:

\[
\boxed{-9384}
\]

Would you like more details or have any questions?

Here are 5 related questions to explore:
1. How does the difference of squares identity help simplify expressions?
2. What is the sum of the first \( N \) natural numbers?
3. Can you derive a formula for the sum of an arithmetic series?
4. How would the sum change if the sequence was extended further?
5. What are the properties of even and odd numbers in such sequences?

**Tip:** When dealing with alternating series, look for patterns or identities that can simplify the expression.

Learn how to solve the alternating squares series problem 1^2 - 2^2 + 3^2 - 4^2 + ... + 135^2 - 136^2 step-by-step. This problem involves sequences, series, and the difference of squares formula. Suitable for advanced high school students and those interested in advanced mathematics.

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