Question

If $x$ is positive, the first negative term in the expansion of $(1+x{)}^{27/5}$ is

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

Answer 1

The correct option is B $8^{\text{th}}$ term

To find first negative term for $(1+x{)}^{27/5}$ consider $|x|>1$ then by taking some constant common are can make it $|x|<1$

So for to solve this question consider $|x|<1$ as this will not change the answer

$(1+x{)}^n=1+nx+\frac{n(n-1)}{1!}{x}^2+.......+\frac{n(n-1)...(n-r+1)}{r!}{x}^r$

$(1+x{)}^{27/5}=1+\frac{27}{5}x+\frac{27/5\left(27/5^{-1}\right)x^2}{1!}+.........$

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

for $T_{r+1}<0$ as $x^r>0$

$n(n-1).....(n-r+1)<0$

$\stackrel{>0}{\frac{27}{5}}(\stackrel{>0}{\frac{27}{5}}-1)(\stackrel{>0}{\frac{27}{5}}-2)(\stackrel{>0}{\frac{27}{5}}-3)(\stackrel{>0}{\frac{27}{5}}-4)(\stackrel{>0}{\frac{27}{5}}-5)(\stackrel{<0}{\frac{27}{5}}-6)<0$

Thus for least value $r$ for whihc $T_{r+1}$ is $<0$ is $7(-r+1=r\Rightarrow r=7)$

$(\stackrel{>0}{\frac{27}{5}}-2)$

Thus $T_8<0$ (first term)

$8^{th}$ term $<0$