Find the 4th term from the beginning and end in the expansion

Question:

Find the $4^{\text {th }}$ term from the beginning and end in the expansion

of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{\mathrm{n}}$

Solution:

To Find :

I. $4^{\text {th }}$ term from the beginning

II. $4^{\text {th }}$ term from the end

Formulae :

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

$\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)$

For $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{\mathrm{n}}$

$a=\sqrt[3]{2}, b=\frac{1}{\sqrt[3]{3}}$ and $n=9$

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

Therefore,

I. For the $4^{\text {th }}$ 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 $t_{4}, r=3$

$\therefore \mathrm{t}_{4}=\mathrm{t}_{3+1}$

$=\left(\begin{array}{l}n \\ 3\end{array}\right)(\sqrt[3]{2})^{n-3}\left(\frac{1}{\sqrt[3]{3}}\right)^{3}$

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

$=\left(\begin{array}{l}n \\ 3\end{array}\right) \cdot \frac{(2)^{\frac{n-3}{3}}}{3}$

$=\frac{n !}{(n-3) ! \times 3 !} \cdot \frac{(2)^{\frac{n-3}{3}}}{3}$

Therefore, $a 4^{\text {th }}$ term from the starting $=\frac{n !}{(n-3) ! \times 3 !} \cdot \frac{(2)^{\frac{n-3}{3}}}{3}$

Now,

II. For the $4^{\text {th }}$ term from the end

We have a formula,

$t_{r+1}=\left(\begin{array}{l}n \\ r\end{array}\right) a^{n-r} b^{r}$

For $t_{(n-2)}, r=(n-2)-1=(n-3)$

$\therefore \mathrm{t}_{(\mathrm{n}-2)}=\mathrm{t}_{(\mathrm{n}-3)+1}$

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

$=\left(\begin{array}{l}\mathrm{n} \\ 3\end{array}\right)(\sqrt[3]{2})^{3}(3)^{\frac{-(\mathrm{n}-3)}{3}} \ldots \ldots\left(\because\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)\right]$

$=\left(\begin{array}{l}\mathrm{n} \\ 4\end{array}\right)(2)(3)^{\frac{3-\mathrm{n}}{3}}$

$=\frac{n !}{(n-4) ! \times 4 !}(2)(3)^{\frac{3-n}{3}}$

Therefore, a $4^{\text {th }}$ term from the end $=\frac{\mathrm{n} !}{(\mathrm{n}-4) ! \times 4 !}(2)(3)^{\frac{3-\mathrm{n}}{3}}$

Conclusion :

1. $4^{\text {th }}$ term from the beginning $=\frac{\mathrm{n} !}{(\mathrm{n}-3) ! \times 3 !} \cdot \frac{(2)^{\frac{\mathrm{n}-3}{3}}}{3}$

II. $4^{\text {th }}$ term from the end $=\frac{\mathrm{n} !}{(\mathrm{n}-4) ! \times 4 !}(2)(3)^{\frac{3-\mathrm{n}}{3}}$