Respuesta :
Answer:
200
Step-by-step explanation:
We have:
[tex]4(-1+2-3+4-5+6-7...+100)[/tex]
We can rearrange the numbers to obtain:
[tex]4((-1-3-5-7-...-99)+(2+4+6...+100))[/tex]
From the left, we can factor out a negative. So:
[tex]4(-(1+3+5+7+...+99)+(2+4+6...+100))[/tex]
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:
[tex](1+3+5+7+...+99)[/tex]
We can use the arithmetic series formula, where:
[tex]S=\frac{k}{2}(a+x_k)[/tex]
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:
[tex]S=\frac{50}{2}(1+99})[/tex]
Divide the fraction. Add within the parentheses:
[tex]S=25(100)[/tex]
Multiply:
[tex]S=2500[/tex]
So, the sum of all the odd terms is 2500.
Even Terms:
We have:
[tex](2+4+6+...+100)[/tex]
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:
[tex]S=\frac{50}{2}(2+100)[/tex]
Divide and add:
[tex]S=25(102)[/tex]
Multiply:
[tex]S=2550[/tex]
We originally had:
[tex]4(-(1+3+5+7+...+99)+(2+4+6...+100))[/tex]
Substitute them for their respective sums:
[tex]4(-(2500)+2550)[/tex]
Multiply:
[tex]4(-2500+2550)[/tex]
Add:
[tex]=4(50)[/tex]
Multiply:
[tex]=200[/tex]
So, the sum of our sequence is 200.
And we're done!
Note: I just found a way easier way to do this. We have:
[tex]4\cdot(-1+2-3+4-5+6-7+...+100)[/tex]
Let's group every two terms together. So:
[tex]=4((-1+2)+(-3+4)+(-5+6)...+(-99+100))[/tex]
We can see that they each sum to 1:
[tex]=4((1)+(1)+(1)+...+(1))[/tex]
Since there are 100 terms, we will have 50 pairs, so 50 times 1. So:
[tex]=4(50)[/tex]
Multiply:
[tex]=200[/tex]
Pick which one you want to use! I will suggest this one though...
Edit: Typo