Answer:
200
Step-by-step explanation:
We have:

We can rearrange the numbers to obtain:

From the left, we can factor out a negative. So:

In other words, we want to find the sum of all the odd numbers from 1 to 99.
And the sum of all the even numbers from 2 to 100.
Let's do each one individually:
Odd Terms:
We have:

We can use the arithmetic series formula, where:

Where k is the number of terms, a is the first term, and x_k is the last term.
Since it's all the odd numbers between 1 and 99, there are 50 terms.
Our first term is 1 and our last term is 99. So, the sum of all the odd terms are:

Divide the fraction. Add within the parentheses:

Multiply:

So, the sum of all the odd terms is 2500.
Even Terms:
We have:

Again, we can use the above formula.
Our first term is 2, last term is 100. And since it's from 2-100, we have 50 even terms. So:

Divide and add:

Multiply:

We originally had:

Substitute them for their respective sums:

Multiply:

Add:

Multiply:

So, the sum of our sequence is 200.
And we're done!
Note: I just found a <em>way</em> easier way to do this. We have:

Let's group every two terms together. So:

We can see that they each sum to 1:

Since there are 100 terms, we will have 50 pairs, so 50 times 1. So:

Multiply:

Pick which one you want to use! I will suggest this one though...
Edit: Typo