Question

Use the method of gauss to find the sum of 1+2+3...400

Answer 1

I just found out what "his method" is. I use a similar one.

I'll refer you to this link where I learned it:
http://answers.yahoo.com/question/index?qid=20080928185405AAKjFcL

Answer 2

\(1\; \; \; +\;\;2 \;\;+\;\;3\;\;+\;\;4\;+\;\;...\;+397+398+399+400\)
\(400+399+398+397+\;...\;+\;\;4\;\;+\;\;3\;+\;\;2\; \;+\;1\)
\(\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\)
\(401+\;401+401+401+\;...\;+401+401+401+401->(400)(401)\)

But, since this is twice our original sum we need to cut it by half, so \(\frac{400(401)}{2}=200(401)\).

I hope this helps.