Let $S_{k} = 3^{k}^{100} C_{0}^{100} C_{k} - 3^{k - 1}^{100} C_{1}^{99} C_{k - 1} + 3^{k - 2}^{100} C_{2}^{98} C_{k - 2} + ... + {(- 1)}^{k}^{100} C_{k}^{100 - k} C_{0},$
then select correct alternative(s) $(k = 0, 1, 2, ....., 100)$

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$S_{66}$ is the greatest amongst these 101 numbers

$S_{1} + S_{3} + S_{5} + .... + S_{99} > S_{2} + S_{4} + .... + S_{100}$

$S_{1} + S_{3} + S_{5} + .... + S_{99} \leq S_{2} + S_{4} + .... + S_{100}$

$S_{67}$ is the greatest amongst these 101 numbers

answer is A, B.

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Detailed Solution

$S_{k} = \sum_{r = 0}^{k} 3^{k - r}^{100} C_{r}^{100 - r} C_{k - r} {(- 1)}^{r}$

$S_{k} = \sum_{r = 0}^{k} 3^{k}^{100} C_{k} (^{k} C_{r} {(- \frac{1}{3})}^{r})$

$S_{k} = 2^{k}^{100} C_{k}$

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