Find the 5th term from the end in the expansion of

Question:

Find the $5^{\text {th }}$ term from the end in the expansion of $\left(x-\frac{1}{x}\right)^{12}$

Solution:

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 $=495(\mathrm{x})^{-4}$

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