Let $x$ is positive, if $k^{th}$ term is the first negative term in the expansion of $(1 + x)^{\frac{31}{5}}, (| x | < 1)$ , then $k =$

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Solution

General term in expansion of $(1 + x)^{n}$ is $T_{r + 1} = \frac{n (n - 1) (n - 2) . . . (n - r + 1)}{r!} x^{r}$
For first negative term,
$n - r + 1 < 0$
$\Rightarrow \frac{31}{5} - r + 1 < 0$
$\Rightarrow r > \frac{36}{5}$
Thus, first negative term occurs when $r = 8$ .
$∴ 9^{th}$ will be first $- ive$ term
Hence $k = 9$

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