We are tasked with finding the sum of the series formed by alternately adding and subtracting consecutive odd integers starting from 1 and ending at 1285, as follows:

$1 - 3 + 5 - 7 + \cdots + 1281 - 1283 + 1285$

Quick Trick for Solving this Problem:

1. Pair the Terms Together:

Notice that for every pair of terms like $(1 - 3), (5 - 7), \cdots, (1281 - 1283)$, each pair results in $-2$: $(1 - 3) = -2, \quad (5 - 7) = -2, \quad (9 - 11) = -2, \quad \dots$

Each pair of consecutive odd numbers sums to $-2$.

2. Count the Pairs:

Odd numbers from 1 to 1285 form an arithmetic sequence. The first odd number is 1, and the last is 1285. The number of odd numbers in this sequence can be found by the formula for the $n$-th odd number, which is $2n - 1$.

To find $n$, solve for: $2n - 1 = 1285 \implies 2n = 1286 \implies n = 643$ So, there are 643 odd numbers in total.

3. Handle the Last Term Separately:

Since we have an odd number of terms, the last term, $1285$, does not have a pair. We sum all the pairs of two consecutive odd numbers, and then add $1285$ at the end.

4. Compute the Sum:

There are $\frac{643 - 1}{2} = 321$ pairs of consecutive odd numbers. Each pair sums to $-2$, so the total sum for the pairs is: $321 \times (-2) = -642$ Now, add the final term $1285$: $-642 + 1285 = 643$

Final Answer:

The sum of the series is $\boxed{643}$.

Would you like a detailed breakdown or more examples?

Here are five related questions for further practice:

1. What is the sum of all odd numbers between 1 and 1000, with alternating signs?
2. How would the sum change if the series were $1 + 3 - 5 + 7 - \cdots + 1285$?
3. Can you derive a general formula for sums of alternating odd sequences ending at any odd number?
4. What if instead of subtracting every second term, you subtract every third term?
5. How many terms would be there in a series from 1 to 2003, and what would be the sum using the same rule?

Tip: Whenever you have alternating sums, always look for easy-to-handle patterns like pairing terms!