Question 285791: Consider the integers from 1 to 100 inclusive. What is the difference between the sum of all the even integers and the sum of all the odd integers?
A. 1 B. 2 C. 0 D. 4 • E. 50
Answer by nabla(475)   (Show Source):
You can put this solution on YOUR website! Let's denote our odds as:
SUM(k=0->49)[2k+1]
and our evens as:
SUM(k=0->50)[2k]
So, we want to evaluate:
SUM(k=0->50)[2k]-SUM(k=0->49)[2k+1]
this is equivalent to:
2*SUM(k=0->50)[k]-2SUM(k=0->49)[k]-50
which is moreover equivalent to:
2*50+2*SUM(k=0->49)[k]-2SUM(k=0->49)[k]-50
Our summation notation cancels, and we are left with 100-50=50. So we shall pick E as the solution.
If these properties of series are unfamiliar, note first that SUM(k=0->49)[1] is 1 added up 50 times (=50). Moreover, When we easily remove first or last members of the series by simply adding the term, hence 2*SUM(k=0->50)[k]=2*50+SUM(k=0->49)[k].
Hope this helps.
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