Find the coefficient of

Solution:

(i) Here, $a=x, b=3$ and $n=8$

We have a formula

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

$=\left(\begin{array}{l}8 \\ r\end{array}\right)(x)^{8-r}(3)^{r}$

$=\left(\begin{array}{l}8 \\ r\end{array}\right)(3)^{r}(x)^{8-r}$

To get coefficient of $x^{5}$ we must have,

$(x)^{8-r}=x^{5}$

- $8-r=5$

- $r=3$

Therefore, coefficient of $x^{5}=\left(\begin{array}{l}8 \\ 3\end{array}\right)(3)^{3}$

$=\frac{8 \times 7 \times 6}{3 \times 2 \times 1} .(27)$

$=1512$

(ii) Here, $a=3 x^{2}, b=\frac{-1}{3 x}$ and $n=9$

We have a formula,

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

$=\left(\begin{array}{l}9 \\ r\end{array}\right)\left(3 x^{2}\right)^{9-r}\left(\frac{-1}{3 x}\right)^{r}$

$=\left(\begin{array}{l}9 \\ r\end{array}\right)(3)^{9-r}\left(x^{2}\right)^{9-r}\left(\frac{-1}{3}\right)^{r}(x)^{-r}$

$=\left(\begin{array}{l}9 \\ r\end{array}\right)(3)^{9-r}(x)^{18-2 r}\left(\frac{-1}{3}\right)^{r}(x)^{-r}$

$=\left(\begin{array}{l}9 \\ \mathrm{r}\end{array}\right)(3)^{9-\mathrm{r}}(\mathrm{x})^{18-2 \mathrm{r}-\mathrm{r}}\left(\frac{-1}{3}\right)^{\mathrm{r}}$

$=\left(\begin{array}{l}9 \\ \mathrm{r}\end{array}\right)(3)^{9-\mathrm{r}}\left(\frac{-1}{3}\right)^{\mathrm{r}}(\mathrm{x})^{18-3 \mathrm{r}}$

To get coefficient of $x^{6}$ we must have,

$(x)^{18-3 r}=x^{6}$

- $18-3 r=6$

- $3 r=12$

- $r=4$

Therefore, coefficient of $x^{6}$ $=\left(\begin{array}{l}9 \\ 4\end{array}\right)(3)^{9-4}\left(\frac{-1}{3}\right)^{4}$

$=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \cdot(3)^{5}\left(\frac{1}{3}\right)^{4}$

$=126 \times 3$

$=378$

(iii) Here, $a=3 x^{2}, b=\frac{-a}{3 x^{3}}$ and $n=10$

We have a formula,

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

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

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

$=\left(\begin{array}{c}10 \\ r\end{array}\right)(3)^{10-r}(x)^{20-2 r}\left(\frac{-a}{3}\right)^{r}(x)^{-3 r}$

$=\left(\begin{array}{c}10 \\ r\end{array}\right)(3)^{10-r}(x)^{20-2 r-3 r}\left(\frac{-a}{3}\right)^{r}$

$=\left(\begin{array}{c}10 \\ r\end{array}\right)(3)^{10-r}\left(\frac{-a}{3}\right)^{r}(x)^{20-5 r}$

To get coefficient of $\mathrm{x}^{-15}$ we must have,

$(x)^{20-5 r}=x^{-15}$

- $20-5 r=-15$

- $5 r=35$

- $r=7$

Therefore, coefficient of $x^{-15}=\left(\begin{array}{c}10 \\ 7\end{array}\right)(3)^{10-7}\left(\frac{-a}{3}\right)^{7}$

But $\left.\left(\begin{array}{c}10 \\ 7\end{array}\right)=\left(\begin{array}{c}10 \\ 3\end{array}\right) \ldots \ldots \ldots\left(\begin{array}{c}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)\right]$

Therefore, the coefficient of $\mathrm{x}^{-15}=\frac{10 \times 9 \times 8}{3 \times 2 \times 1} \cdot(3)^{3}\left(\frac{-\mathrm{a}}{3}\right)^{7}$

$=120 \cdot(-a)^{7}\left(\frac{1}{3}\right)^{4}$

$=(-a)^{7} \frac{120}{3^{4}}$

$=(-a)^{7} \frac{40}{27}$

(iv) Here, $a=a, b=-2 b$ and $n=12$

We have formula

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

$=\left(\begin{array}{c}12 \\ r\end{array}\right)(a)^{12-r}(-2 b)^{r}$

$=\left(\begin{array}{c}12 \\ r\end{array}\right)(-2)^{r}(a)^{12-r}(b)^{r}$

To get coefficient of $a^{7} b^{5}$ we must have,

$(a)^{12-r}(b)^{r}=a^{7} b^{5}$

$\cdot r=5$

Therefore, coefficient of $a^{7} b^{5}=\left(\begin{array}{c}12 \\ 5\end{array}\right)(-2)^{5}$

$=\frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} \cdot(-32)$

$=792 .(-32)$

$=-25344$