What is the sum of all even numbers from 0-200?

Question

you know like, 2+4+6+8..... all the way to 200

Answer 1

example:
a) Four first even number:
2 + 4 + 6 + 8 = 20 = 4 * (4+1) = 4 * 5
b) Eight first even number:
2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 = 72 = 8 * (8+1) = 8 * 9

so, the answer to your question is
100*101=10,100

Answer 2

10100

Rearrange the numbers by pairing the first number with the last, the second number with the second from last, the third number with the third from last, and so on like this:

(2 + 200) + (4 + 198) + (6 + 196) + ..... + (98 + 104) + (100 + 102)

Notice that the sum of each pair is 202, and there are 50 pairs. So the sum of all the numbers is 202 X 50 = 10100

Answer 3

Pair the even numbers from the outside to the inside like so

2+200 = 202
4+198 = 202
6+196 = 202
...
...
98+104 = 202
100+102 = 202

Note that there are 50 pairs so

202 * 50 = 10,100

Answer 4

Another way to solve this problem would be to look at it as an arithmetic series. To find the sum of an arithmetic series, you will need to know three things: the first term, the last term, and the number of terms. You stated in your original question that the first and last numbers are 2 and 200, respectively. Obviously, the number of terms you will have is 100 (divide each term by two if you don't see this).

Back to finding the sum, the sum of an arithmetic series is the following:

S = [number of terms * (first term + last term)] / 2
S = [100 * (2+200)] / 2
S = [100 * 202] / 2
S = 10100

And your final answer is, 10100.

Going one step further if you so desire:

For an abbreviated way of writing this series, we can write it in sigma notation. http://img95.imageshack.us/img95/1421/sigmagl5.gif

What that image, translated into English, is "the sum of the series (2x) as x goes from 1 to 100". Basically, you'll be putting each integer between 1 and 100 in for x, and you get your original series: 2,4,6,...,196,198,200

Answer 5

Add 1st 0 + 200 = 200
Add 2nd 2 + 198 = 200
Add 3rd 4 + 196 = 200
Add 4th 6 + 194 = 200
Add 5th 8 + 192 = 200
....
Add 50th 98 + 102 = 200
Add 51th 100 +100 = 200
So it would be 51 * 200 = 10,200.

Answer 6

you could pair the numbers and add them:
200+0=200
199+1=200
198+2=200...
99+101=200
but that takes too long but you can still use this idea: so , 200*100+100=20100...you have to add 100 because it wasn't part of the paring

Answer 7

--- 2+4+6+8+...+198+200
--- 2(1+2+3+4+...+99+100)
--- 2[ (1+100)+(2+99)+(3+98)+... +(50+51) ]
--- 2(101+101+101+101+101 ... +101)
--- 2(101*50)
--- 100*101
--- 10100

Answer 8

The proper answer is 10100
If you pair all the even numbers you et something like this:
2+200,4+198, etc
this gives you 202 times the number of combinations.
from (halfway), there are 50 pairs
So: 50 times 202 = 10100

Answer 9

100

Answer 10

10,100