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If $\mathrm{x}$ is positive, the first negative term in the expansion of $(1+x)^{27 / 5}$ is
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The correct answer is:
8th term
8th term
$T_{r+1}=\frac{n(n-1)(n-2) \ldots \ldots \ldots(n-r+1)}{r !}(x)^r$
For first negative term, $n-r+1 < 0$ or $r>\frac{32}{5}$
$\therefore \mathrm{r}=7$. Therefore, first negative term is $\mathrm{T}_8$.
For first negative term, $n-r+1 < 0$ or $r>\frac{32}{5}$
$\therefore \mathrm{r}=7$. Therefore, first negative term is $\mathrm{T}_8$.
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