Discussion
Find the sum of
\[\begin{aligned}&\left(1-\tfrac{1}{99}\right)+\left(1-\tfrac{2}{99}\right)\\&+\left(1-\tfrac{3}{99}\right)+\cdots+\left(1-\tfrac{98}{99}\right)\end{aligned}\]Answer: B
The sum is 49. There are 98 terms, each of the form 1 minus k/99. Their sum is 98 minus (1 + 2 + ... + 98)/99. The sum of 1 to 98 is 98 x 99 / 2 = 4851, and 4851/99 = 49. So the total is 98 - 49 = 49. Option 98 forgets to subtract the fractions, and 99 and 50 do not match either the term count or the subtracted total. As a check, pair the first and the last terms: (1 - 1/99) + (1 - 98/99) = 2 - 99/99 = 1. There are 49 such pairs among the 98 terms, so the sum is 49 x 1 = 49, agreeing with the main working.
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