Question 1094794
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A = 20000*(1+2+3+...+20001) = {{{20000*((20001*20002)/2)}}} = {{{(1/2)*20000*20001*20002}}}.


B = 20001*(1+2+3+...+20000) = {{{20001*((20000*20001)/2)}}} = {{{(1/2)*20000*20001^2}}}.



Now, comparing these expressions, you may conclude (without evaluating) that  A > B.



<U>HINT</U>.  I used well known formula for the sum of the first "n" natural numbers


          1 + 2 + 3 + . . . + n = {{{(n*(n+1))/2}}}.
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