Question

If ${}^{\prime }{x}^{\prime }$ is positive; the first negative terms in the expansion of $(1+x{)}^{\frac{27}{5}}(|x|<1)$ is

• A. $7^{th}$ term
• B. $5^{th}$ term
• C. $8^{th}$ term
• D. $6^{th}$ term

Answer 1

The correct option is C $8^{th}$ term

Solution:- (C) $8^{th}$ term

In the expansion of ${(1+x)}^{\frac{27}{5}}$, coefficient of ${(r+1)}^{th}$ term-

$T_{r+1}=\frac{\left(\frac{27}{5}\right)(\frac{27}{5}-1).....(\frac{27}{5}-r+1)}{r!}{x}^r$

$T_{r+1}<0$ only if the last term of the numerator is less than zero because all other term and denominator are positive.

$\therefore \frac{27}{5}-r+1<0$

$\Rightarrow r>\frac{32}{5}$

Hence smallest vaue of $r$ for which $T_{r+1}<0$ is $7$.

$\therefore$ Negative term will be $T_{7+1}=T_8$

Hence $8^{th}$ term of the given expression will be negative.