document.write( "Question 1142839: how do you add 1+ 2+ 3+ +4 all the way up to say 200 or 300? Is there an equation to use for that? \n" ); document.write( "

Algebra.Com's Answer #763569 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "Pick a small number for an example that you can see easily:

\n" ); document.write( "1+2+3+4+5+6+7+8+9+10 = ??

\n" ); document.write( "Group the numbers in pairs, like this:

\n" ); document.write( "(1+10)+(2+9)+(3+8)+(4+7)+(5+6)

\n" ); document.write( "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.

\n" ); document.write( "It works when the last number is odd also. For example,

\n" ); document.write( "1+2+3+4+5+6+7 = (1+7)+(2+6)+(3+5)+4

\n" ); document.write( "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.

\n" ); document.write( "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:

\n" ); document.write( " $\"%28n%2F2%29%28n%2B1%29\"$

\n" ); document.write( "So for your two particular ending numbers,

\n" ); document.write( "1+2+3+...+200 = (200/2)(200+1) = 100*201 = 20100
\n" ); document.write( "1+2+3+...+300 = (300/2)(300+1) = 150*301 = 45150

\n" ); document.write( "Or, for a random odd last number,

\n" ); document.write( "1+2+3+...+79 = (79/2)(79+1) = (79/2)(80) = 79*40 = 3160 \n" ); document.write( "

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