Question

Find the general term in the expansion of${(1-x)}^{-3}$.

Answer 1

Expand the given expression to find the general term

Given, ${(1-x)}^{-3}$

${(r+1)}^{th}$ term in the expansion of ${(1-x)}^n$ is given by:

$$\begin{gathered} \Rightarrow {\mathrm{T}}_{\mathrm{r}+1}=\frac{\mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right).....\left(\mathrm{n}-\mathrm{r}+1\right)}{\mathrm{r}!} \\ \Rightarrow {\mathrm{T}}_{\mathrm{r}+1}=\frac{-3\left(-3-1\right)\left(-3-2\right)......\left(-3-\mathrm{r}+1\right)}{\mathrm{r}!}{\left(-\mathrm{x}\right)}^{\mathrm{r}}[\mathrm{put}n=-3] \\ \Rightarrow {\mathrm{T}}_{\mathrm{r}+1}=\frac{-3\left(-4\right)\left(-5\right)\left(-6\right)......\left(-2-\mathrm{r}\right)}{\mathrm{r}!}{\left(-1\right)}^{\mathrm{r}}{\left(\mathrm{x}\right)}^{\mathrm{r}} \end{gathered}$$

$$\begin{gathered} \Rightarrow {\mathrm{T}}_{\mathrm{r}+1=}{\left(-1\right)}^{\mathrm{r}}{\left(-1\right)}^{\mathrm{r}}\frac{3\left(4\right)\left(5\right)\left(6\right)......\left(2+\mathrm{r}\right)}{1·2·3·4.....\left(\mathrm{r}-1\right)·\mathrm{r}}{\left(\mathrm{x}\right)}^{\mathrm{r}} \\ \Rightarrow {\mathrm{T}}_{\mathrm{r}+1=}\frac{\left(\mathrm{r}+1\right)\left(\mathrm{r}+2\right)}{1·2}{\left(\mathrm{x}\right)}^{\mathrm{r}} \end{gathered}$$ [$\because {(-1)}^r\times {\left(-1\right)}^r=1$]

Hence, the ${r+1}^{th}$term is${\mathrm{T}}_{\mathrm{r}+1=}\frac{\left(\mathrm{r}+1\right)\left(\mathrm{r}+2\right)}{1·2}{\left(\mathrm{x}\right)}^{\mathrm{r}}$.