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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x^2})^m$,where $x \neq 0$ and $m$ is a natural number,is $559$. Find the term of the expansion containing $x^3$. (in $x^3$)

If $C_j$ stands for ${}^nC_j$,then $\frac{C_1}{C_0} + \frac{2 \times C_2}{C_1} + \frac{3 \times C_3}{C_2} + \ldots + \frac{n \times C_n}{C_{n-1}} = $

If the coefficient of $x^3$ in the binomial expansion of $x^3(2 \sqrt{3} x^2 + \frac{1}{kx})^{12}$ is $880$,then $k$ is equal to

The coefficients of the $(r-1)^{th}$,$r^{th}$,and $(r+1)^{th}$ terms in the expansion of $(x+1)^{n}$ are in the ratio $1:3:5$. Find $n$ and $r$.

The sum of the coefficients of the integral powers of $x$ in the binomial expansion of $(1 - 2\sqrt{x})^{50}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module