2018 AMC 12B Problem 9Problem 9 of 25·Easier·AlgebraWhat is ∑i=1100∑j=1100(i+j)? \sum_{i=1}^{100}\sum_{j=1}^{100}(i+j)? i=1∑100j=1∑100(i+j)?Answer choicesA100,100100{,}100100,1001B500,500500{,}500500,5002C505,000505{,}000505,0003D1,001,0001{,}001{,}0001,001,0004E1,010,0001{,}010{,}0001,010,0005Submit answerStuck? Show hintsShow solutionSolutionSplitting the sum, ∑i=1100∑j=1100(i+j)=∑i=1100∑j=1100i+∑i=1100∑j=1100j=100∑i=1100i+100∑j=1100j. \begin{gathered} \sum_{i=1}^{100}\sum_{j=1}^{100}(i+j) \\ =\sum_{i=1}^{100}\sum_{j=1}^{100}i \\ {}+\sum_{i=1}^{100}\sum_{j=1}^{100}j \\ =100\sum_{i=1}^{100}i \\ {}+100\sum_{j=1}^{100}j. \end{gathered} i=1∑100j=1∑100(i+j)=i=1∑100j=1∑100i+i=1∑100j=1∑100j=100i=1∑100i+100j=1∑100j. Since ∑k=1100k=5050,\sum_{k=1}^{100}k=5050,∑k=1100k=5050, this equals 100⋅5050100\cdot5050100⋅5050 +100⋅5050+100\cdot5050+100⋅5050 =1,010,000.=1{,}010{,}000.=1,010,000. Thus, the correct answer is E.AoPS wikiCopy problemTagged: summation · arithmetic sequence