Question 285791
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.