Question

Find the coefficient of the term which contains the 10th power of x in the expansion of ${(x^{\frac{1}{3}}+\frac{1}{x^3\sqrt{x}})}^{40}$.

• A. 870
• B. 680
• C. 780
• D. none of these

Answer 1

The correct option is D none of these

Given, ${(x^{\frac{1}{3}}+\frac{1}{x^3\sqrt{x}})}^{40}$

General term, $T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

$T_{r+1}={}^{40}{C}_r(x^{\frac{1}{3}}{)}^{40-r}{\left(\frac{1}{x^3\sqrt{x}}\right)}^r$

$T_{r+1}={}^{40}{C}_r(x{)}^{\frac{40}{3}-\frac{r}{3}}{\left(\frac{1}{x^{\frac{7}{2}}}\right)}^r$

$T_{r+1}={}^{40}{C}_r(x{)}^{\frac{40}{3}-\frac{23r}{6}}$

We need to find the term which contains ${10}^{th}$ power of x

So, $\frac{40}{3}-\frac{23r}{6}=10$

$\frac{80-23r}{6}=10$

$80-23r=60$

$20=23r$

$r=\frac{20}{23}$

Which is not possible.

Hence, correct option is D.