Let S_n = sum_{k=1}^n k denote the sum of the | vedclass

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(C) We have $S_n = \frac{n(n+1)}{2}$.The sequence of sums is $S_1=1, S_2=3, S_3=6, S_4=10, S_5=15, S_6=21, S_7=28, S_8=36, \ldots$.Observing the parit

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$\frac{49}{99}$

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(C) We have $S_n = \frac{n(n+1)}{2}$.The sequence of sums is $S_1=1, S_2=3, S_3=6, S_4=10, S_5=15, S_6=21, S_7=28, S_8=36, \ldots$.Observing the parity of $S_n$:$S_1$ (odd),$S_2$ (odd),$S_3$ (even),$S_4$ (even),$S_5$ (odd),$S_6$ (odd),$S_7$ (even),$S_8$ (even),...The pattern of parity repeats every $4$ terms: (odd,odd,even,even).In the set ${S_1, S_2, \ldots, S_{99}}$,there are $99$ terms.Since $99 = 4 	imes 24 + 3$,we have $24$ full cycles of (odd,odd,even,even) and the first $3$ terms of the next cycle (odd,odd,even).Number of even terms = $24 	imes 2 + 1 = 49$.Total number of cards = $99$.Probability of drawing an even number = $\frac{	ext{Number of even terms}}{	ext{Total number of terms}} = \frac{49}{99}$.

Let $S_n = \sum_{k=1}^n k$ denote the sum of the first $n$ positive integers. The numbers $S_1, S_2, S_3, \ldots, S_{99}$ are written on $99$ cards. The probability of drawing a card with an even number written on it is

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