Question

If $x$ is positive, the first negative term in the expansions of $(1+x{)}^{\frac{27}{5}}$ is:

• A. $7th$ terms
• B. $5th$ terms
• C. $8th$ terms
• D. $6th$ terms

Answer 1

Answer: (C)

$8th$ terms

The expansion of $(1+x{)}^{\frac{27}{5}}$ coefficient of ${(r+1)}^{th}$ term

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

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

so $\frac{27}{5}-r+1<0\Rightarrow r>\frac{32}{5}=6\frac{2}{5}$

smallest $r$ for which $T_{r+1}<0$ is $r=7$

$T_{7+1}=T_8=8^{th}$ term is negative

Answer C