Question

# Show that the expansion of $$\left( x^2 + \dfrac{1}{x} \right)^{12}$$ does not contain any term involving $$x^{-1}$$.

Solution

## General term is $$T_{r} = ^{n}C_{r}\,x^{r}\,a^{n-r}$$in $$(x+a)r_{n}(r\epsilon z)$$$$\therefore T_{r} = ^{12}C_{r}(x^{2})^{r}\left ( \dfrac{1}{x} \right )^{12-r}$$$$\Rightarrow T_{r} = ^{12}C_{r}\,x^{3r-12}$$for $$x^{-1}$$ term $$3r-12 = -1$$$$\Rightarrow \boxed{r = \dfrac{11}{3}}$$ not possibleMathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More