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For any sequence of n consecutive positive integers, Se denotes the sum of all even integers and So denotes the sum of all odd integers. Which of the following must be true?
1. There is at least one such sequence for which Se > So
2. There is at least one such sequence for which Se = So
3. There is at least one such sequence for which Se < So.
A. 1 only
B. 2 only
C. 3 only
D. 1 & 2 only
E. 1 & 3 only.
30 sec approach:It's easy to get that 1 and 3 must be true. Just consider two easiest sets:
{1, 2} the sum of even integers, which is 2, is
more than the sum of odd integers, which is 1;
{2, 3} the sum of even integers, which is 2, is
less than the sum of odd integers, which is 3;
Since only answer choice E offers both options (1 and 3) then it must be a correct answer. So we don't even need to consider 2.
Answer: E.
Just to elaborate on 3.
There is at least one such sequence for which Se = So: first of all, the sum of even integers is always even, hence the sum of odd integers must also be even, so # of odd terms must be even. Now, consider the set {a, a+1, a+2, a+3} (it really doesn't matter how many terms we choose, since # of odd terms is even and it really doesn't matter whether a is even or odd). In order Se = So to be true the following must be true: a+(a+2)=(a+1)+(a+3) --> 2=4, which is not true, so Se = So is not possible.
Hope it's clear.