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

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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 $t_1+t_2+\ldots+t_k=$

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