<PRE><B>Question 1142839</B>
&lt;br&gt;
Pick a small number for an example that you can see easily:&lt;br&gt;
1+2+3+4+5+6+7+8+9+10 = ??&lt;br&gt;
Group the numbers in pairs, like this:&lt;br&gt;
(1+10)+(2+9)+(3+8)+(4+7)+(5+6)&lt;br&gt;
The sum is now the sum of 5 pairs, with the sum of each pair being 11.  So the sum of all the numbers is 5*11 = 55.&lt;br&gt;
It works when the last number is odd also.  For example,&lt;br&gt;
1+2+3+4+5+6+7 = (1+7)+(2+6)+(3+5)+4&lt;br&gt;
which gives you 3 1/2 pairs, each with a sum of 8; the sum is 8 times 3 1/2 = 4 times 7 = 28.&lt;br&gt;
In general, if the last number is n, the sum of the first and last numbers is 1+n, and the number of pairs is n/2.  That gives you the following general formula for the sum of the integers from 1 to n:&lt;br&gt;
{{{(n/2)(n+1)}}}&lt;br&gt;
So for your two particular ending numbers,&lt;br&gt;
1+2+3+...+200 = (200/2)(200+1) = 100*201 = 20100
1+2+3+...+300 = (300/2)(300+1) = 150*301 = 45150&lt;br&gt;
Or, for a random odd last number,&lt;br&gt;
1+2+3+...+79 = (79/2)(79+1) = (79/2)(80) = 79*40 = 3160</PRE>