Problem:
What is
i=1∑100j=1∑100(i+j)?
Answer Choices:
A. 100,100
B. 500,500
C. 505,000
D. 1,001,000
E. 1,010,000
Solution:
Note that the sum of the first 100 positive integers is 21⋅100⋅101=5050. Then
i=1∑100j=1∑100(i+j)=i=1∑100j=1∑100i+i=1∑100j=1∑100j=j=1∑100i=1∑100i+i=1∑100j=1∑100j
=100i=1∑100i+100j=1∑100j=100(5050+5050)=1,010,000
OR
Note that the sum of the first 100 positive integers is 21⋅100⋅101= 5050. Then
i=1∑100j=1∑100(i+j)=i=1∑100((i+1)+(i+2)+⋯+(i+100))=i=1∑100(100i+5050)=100⋅5050+100⋅5050=1,010,000
OR
The sum contains 10,000 terms, and the average value of both i and j is 2101, so the sum is equal to
10,000(2101+2101)=1,010,000
The problems on this page are the property of the MAA's American Mathematics Competitions