Question

In the expansion of $\left(x^{\frac{4}{5}} + x^{- \frac{1}{5}}\right)^{n}$ , the coeficient of the $8^{t h}$ and $19^{t h}$ terms are equal. The term independent of $x$ is given by

• A. $\_{}^{27}C_{21}^{}$
• B. $\_{}^{25}C_{20}^{}$
• C. $\_{}^{25}C_{21}^{}$
• D. $\_{}^{27}C_{22}^{}$

Answer 1

Answer: (B)

Since there is no constant term the coefficient of $8^{\text{th}}$ and $19^{\text{th}}$ term are same as the binomial coefficients of $8^{\text{th}}$ and $19^{\text{th}}$ term.
$\_{}^{n}C_{7}^{}=^{n}C_{18}\Rightarrow n=7+18=25$
$T_{n + 1}=^{25}C_{r}\left(x^{\frac{4}{5}}\right)^{25 - r}\left(r^{- \frac{1}{5}}\right)^{r}$
$\_{}^{25}C_{r}^{} \, x^{\frac{4}{5} \left(\right. 25 - r \left.\right) - \frac{r}{5}}$
To be independent of $x,\frac{100 - 5 r}{5}=0\Rightarrow r=20$
Hence, the required term is $\_{}^{25}C_{20}^{}$