Question

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

Answer 1

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 $r=8$.
$\therefore 9^{\text{th}}$ will be first negative term
Hence $k=9$