### Binomial Theorem Question 19

#### Question 19 - 31 January - Shift 2

If the constant term in the binomial expansion of $(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell}})^{9}$ is -84 and the Coefficient of $x^{-3 \ell}$ is $2^{\alpha} \beta$, where $\beta<0$ is an odd number, Then $|\alpha \ell-\beta|$ is equal to_________

## Show Answer

#### Answer: 98

#### Solution:

#### Formula: General term in the expansion

General term of $(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell}})^{9}$ is given by

$T _{r+1}={ }^{9} C_r \frac{(x^{5 / 2})^{9-r}}{2^{9-r}}(\frac{-4}{x^{\ell}})^{r}$

$=(-1)^{r} \frac{{ }^{9} C_r}{2^{9-r}} 4^{r} x^{\frac{455 r}{2-2}}$

$=45-5 r-21 r=0$

$r=\frac{45}{5+21}$

Now, according to the question,

$(-1)^{r} \frac{{ }^{9} C_r}{2^{9-r}} 4^{r}=-84$

$(-1)^{r}{ }^{9} C_r 2^{3 r-9}=21 \times 4$

Only natural value of $r$ possible if $3 r-9=0$

$r=3$ and ${ }^{9} C_3=84$

$\therefore l=5$ from equation (1)

Now, coefficient of $x^{-31}=x^{\frac{45}{2}-\frac{5 r}{2}-1 x}$ at $l=5$, gives

$r=5$

$\therefore{ }^{9} c_5(-1) \frac{4^{5}}{2^{4}}=2^{\alpha} \times \beta$

$2^{\alpha} \times \beta=-63 \times 2^{7}$

$\Rightarrow \alpha=7, \beta=-63$

$\therefore$ value of $|\alpha \ell-\beta|=98$