The positive integer k for which frac{(101)^{ | vedclass

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(B) Let $f(k) = \frac{(\sqrt{101})^k}{k!}$.To find the maximum value,we examine the ratio $\frac{f(k)}{f(k-1)} = \frac{(\sqrt{101})^k}{k!} 	imes \fra

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) Let $f(k) = \frac{(\sqrt{101})^k}{k!}$.To find the maximum value,we examine the ratio $\frac{f(k)}{f(k-1)} = \frac{(\sqrt{101})^k}{k!} 	imes \frac{(k-1)!}{(\sqrt{101})^{k-1}} = \frac{\sqrt{101}}{k}$.We want to find $k$ such that $f(k) \ge f(k-1)$,which implies $\frac{\sqrt{101}}{k} \ge 1$,or $k \le \sqrt{101}$.Since $\sqrt{101} \approx 10.05$,the condition $k \le 10.05$ holds for $k = 1, 2, \dots, 10$.This means $f(1) For $k > 10$,the ratio $\frac{\sqrt{101}}{k} Thus,the sequence increases until $k=10$ and then decreases.Therefore,the maximum value occurs at $k = 10$.

The positive integer $k$ for which $\frac{(101)^{k/2}}{k!}$ is a maximum is

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