Question
If $$x$$ is positive, the first negative term in the expansion of $${\left( {1 + x} \right)^{\frac{{27}}{5}}}$$ is
A.
$${6^{th}}$$ term
B.
$${7^{th}}$$ term
C.
$${5^{th}}$$ term
D.
$${8^{th}}$$ term
Answer :
$${8^{th}}$$ term
Solution :
$${T_{r + 1}} = \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)......\left( {n - r + 1} \right)}}{{r!}}{\left( x \right)^r}$$
For first negative term, $$n - r + 1 < 0$$
$$ \Rightarrow \,\,r > n + 1$$
$$ \Rightarrow \,\,r > \frac{{32}}{5}\,\,\,\,\,\therefore \,\,r = 7.\left( {\because \,\,n = \frac{{27}}{5}} \right)$$
Therefore, first negative term is $${T_8}.$$