Find the 6th term of the expansion

Solution:

Given : $3^{\text {rd }}$ term from the end $=45$

To Find : $6^{\text {th }}$ term

For $\left(y^{1 / 2}+x^{1 / 3}\right)^{n}$

$a=y^{1 / 2}, b=x^{1 / 3}$

We have $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

As $n=n$, therefore there will be total $(n+1)$ terms in the expansion.

$3^{\text {rd }}$ term from the end $=(n+1-3+1)^{\text {th }}$ i.e. $(n-1)^{\text {th }}$ term from the starting

For $(n-1)^{\text {th }}$ term, $r=(n-1-1)=(n-2)$

$t_{(n-1)}=t_{(n-2)+1}$

$=\left(\begin{array}{c}n \\ n-2\end{array}\right)\left(y^{\frac{1}{2}}\right)^{n-(n-2)}\left(x^{\frac{1}{3}}\right)^{(n-2)}$

$=\left(\begin{array}{l}\mathrm{n} \\ 2\end{array}\right)\left(y^{\frac{1}{2}}\right)^{2}(x)^{\frac{\mathrm{n}-2}{3}}$ ............$\because\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)$

$=\frac{\mathrm{n}(\mathrm{n}-1)}{2}(\mathrm{y})(\mathrm{x})^{\frac{\mathrm{n}-2}{3}}$

Therefore $3^{\text {rd }}$ term from the end $=\frac{n(n-1)}{2}(y)(x)^{\frac{n-2}{3}}$

Therefore coefficient $3^{\text {rd }}$ term from the end $=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$

$\therefore 45=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$

- $90=n(n-1)$

- $10(9)=n(n-1)$

Comparing both sides, n=10

For $6^{\text {th }}$ term, $r=5$

$t_{6}=t_{5+1}$

$=\left(\begin{array}{c}10 \\ 5\end{array}\right)\left(y^{\frac{1}{2}}\right)^{10-5}\left(x^{\frac{1}{3}}\right)^{5}$

$=\left(\begin{array}{c}10 \\ 5\end{array}\right)(\mathrm{y})^{\frac{5}{2}}(\mathrm{x})^{\frac{5}{3}}$

$=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}(\mathrm{y})^{\frac{5}{2}}(\mathrm{x})^{\frac{5}{3}}$

$=252(\mathrm{y})^{\frac{5}{2}}(\mathrm{x})^{\frac{5}{3}}$

$\underline{\text { Conclusion }}: 6^{\text {th }}$ term $=252(\mathrm{y})^{\frac{5}{2}}(\mathrm{x})^{\frac{5}{3}}$