For x=frac{5}{7},if t_k is the first negative | vedclass

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

Question

(C) The expansion of $(1+x)^n$ for $|x| < 1$ is given by $1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots$
Given $n = \frac{7}{

Vedclass · Accepted Answer

$\frac{104}{49}$

Vedclass · Answer

(C) The expansion of $(1+x)^n$ for $|x| < 1$ is given by $1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots$
Given $n = \frac{7}{5}$ and $x = \frac{5}{7}$.
$t_1 = 1$
$t_2 = nx = \frac{7}{5} 	imes \frac{5}{7} = 1$
$t_3 = \frac{n(n-1)}{2!}x^2 = \frac{\frac{7}{5}(\frac{2}{5})}{2} 	imes (\frac{5}{7})^2 = \frac{7}{25} 	imes \frac{25}{49} = \frac{1}{7}$
$t_4 = \frac{n(n-1)(n-2)}{3!}x^3 = \frac{\frac{7}{5}(\frac{2}{5})(-\frac{3}{5})}{6} 	imes (\frac{5}{7})^3 = \frac{-\frac{42}{125}}{6} 	imes \frac{125}{343} = -\frac{7}{343} = -\frac{1}{49}$
Since $t_4$ is the first negative term,$k=4$.
The sum is $t_1+t_2+t_3+t_4 = 1 + 1 + \frac{1}{7} - \frac{1}{49} = 2 + \frac{7-1}{49} = 2 + \frac{6}{49} = \frac{98+6}{49} = \frac{104}{49}$.

For $x=\frac{5}{7}$,if $t_k$ is the first negative term in the expansion of $(1+x)^{7/5}$,then $t_1+t_2+\ldots+t_k=$

Similar Questions

If $\frac{(1 - 3x)^{1/2} + (1 - x)^{5/3}}{\sqrt{4 - x}}$ is approximately equal to $a + bx$ for small values of $x$,then $(a,b) = $

If the $(r + 1)^{th}$ term is the first negative term in the expansion of $(1 + x)^{7/2}$,then the value of $r$ is

If $|x| < 1$,then the coefficient of $x^n$ in the expansion of $(1 + x + x^2 + ....)^2$ will be

The coefficient of $x^2$ in the expansion of $(1+x)^2(8-x)^{-\frac{1}{3}}$ is

The expansion of $\frac{1}{\sqrt{4 - 3x}}$ using the binomial theorem is valid if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module