Find the $5^{\text {th }}$ term from the end in the expansion of $\left(x-\frac{1}{x}\right)^{12}$
To Find : $5^{\text {th }}$ term from the end
Formulae :
- $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$
$\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)$
For $\left(x-\frac{1}{x}\right)^{12}$
$\mathrm{a}=\mathrm{x}, \quad \mathrm{b}=\frac{-1}{\mathrm{x}}$ and $\mathrm{n}=12$
As n=12 , therefore there will be total (12+1)=13 terms in the expansion
Therefore,
$5^{\text {th }}$ term from the end $=(13-5+1)^{\text {th }}$ i.e. $9^{\text {th }}$ term from the starting.
We have a formula
$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$
For tg, $r=8$
$\therefore \mathrm{t}_{9}=\mathrm{t}_{8+1}$
$=\left(\begin{array}{c}12 \\ 8\end{array}\right)(\mathrm{x})^{12-8}\left(\frac{-1}{\mathrm{x}}\right)^{8}$
$=\left(\begin{array}{c}12 \\ 4\end{array}\right)(\mathrm{x})^{4}(\mathrm{x})^{-8} \ldots \ldots\left[\because\left(\begin{array}{c}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)\right]$
$=\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}(\mathrm{x})^{4-8}$
$=495(\mathrm{x})^{-4}$
Therefore, a $5^{\text {th }}$ term from the end $=495(\mathrm{x})^{-4}$
$\underline{\text { Conclusion : }} 5^{\text {th }}$ term from the end $=495(\mathrm{x})^{-4}$