  <p>I understand that there is a way to get this answer by simply (yet time comsumingly) doing the math or trying to figure out a pattern on the spot. I have started to see numerous questions like this one, however, and am pretty sure that there's an easier way to approach this kind of problem i just don't know what it is :)...any insights? Thank you!!</p>

<p>The sum of the positive odd integers less than 100 is subtracted from the sum of the positive even integers less 100 than or equal to 100 . What is the resulting difference?</p>

<p>This kind of problem does come up often. The idea is to think of how to group the calculations to save time. In this case, there are 50 even numbers you are adding --</p>

<p>2, 4, 6, ...., 98, 100</p>

<p>and then 50 odd #s --</p>

<p>1, 3, 5 ,..., 97, 99</p>

<p>But notice that you can pair each even with the corresponding odd, take the difference and keep a running total. So you'd have: (2-1) + (4-3) + (6-5) +...+ (98-97) + (100 - 99)</p>

<p>Each pair has a difference of 1 and there are 50 of these pairs, so the answer is 50.</p>

<p>I'm trying to know if there's an easier way to do this too.</p>

<p>Or when you get questions like: "What is the sum of the integers from 0 to 200 inclusively?"</p>

<p>Odd and even numbers are off by 1 (1 and 2; 3 and 4; 4 and 5; 5 and 6; etc.) There are 50 odd numbers and 50 even numbers in the set <a href="inclusive%20of%200%20and%20100">0, 100</a>. Thus, the difference between the sums of the two sets is (1)*(50) = 50. If you don't understand this then you'll have to list the numbers out and see for yourself.</p>

<p>
[quote]
What is the sum of the integers from 0 to 200 inclusively?

[/quote]

The average of the numbers in the set [0, 200] is 100. You multiply the average by the number of numbers in the set: 100*201 = 20100</p>

<h1>0 + 1 + 2 + 3 + 4 + 5 + . . . + 100 + . . . + 195 + 196 + 197 + 198 + 199 + 200</h1>

<h1>(100-100) + (100-99) + (100-98) + (100-97) + (100-96) + (100-95) + . . . + 100 + . . . + (100+95) + (100+96) + (100+97) + (100+98) + (100+99) + (100+100)</h1>

<p>100<em>(how many numbers there are in the set)
100</em>(201) = 20100</p>

<p>There is a much simpler method but it involves PreCalculus. It is as follows:</p>

<p>1.) Write the sequence:
0,1,2,3,4....200</p>

<p>2.) The formula for an arithmetic series is y=first term + (n-1)difference between two consecutive terms ===> y=0+(n-1)1
y=n-1
3.)Find number of terms by substituting last term for y
200=n-1
n=201
4.) Use sum of finite arithmetic sequence formula: S= n(first term + last term)/2
S=201(0+200)/2
S=20,100</p>