If the constant term in the binomial expansion of (x^(5)/(2)2-4x^l)^9 is -84 and the coeff…

Question

If the constant term in the binomial expansion of ${(\frac{{x}^{\frac{5}{2}}}{2}-\frac{4}{{x}^{l}})}^{9}$ is $-84$ and the coefficient of ${x}^{-3l}$ is

${2}^{\alpha }\beta $, where $\beta <0$ is an odd number, then $|\alpha l-\beta |$ is equal to \_\_\_\_\_\_\_\_.

Answer: 98

Accepted Answer

Given binomial expansion of ${(\frac{{x}^{\frac{5}{2}}}{2}-\frac{4}{{x}^{l}})}^{9}$

$\begin{gathered} {T}_{r+1} ={}^{9}{C}_{r}{(\frac{{x}^{\frac{5}{2}}}{2})}^{9-r}{(\frac{-4}{{x}^{l}})}^{r} \ ={}^{9}{C}_{r}{x}^{\frac{45-5r}{2}-lr}\cdot {2}^{r-9}\cdot {4}^{r}\cdot (-1{)}^{r} \end{gathered}$

For constant term, power of x is zero.

So, $\frac{45-5r}{2}=lr\Rightarrow 2lr+5r=45$

Now constant term $=-84$

and ${}^{9}{C}_{r}\cdot {2}^{3r-9}(-1{)}^{r}=-84$

So, $r=3$ and $l=5$

Now for ${x}^{-15},\frac{45-5r}{2}-5r=-15$

$\Rightarrow$ $45-15r=-30$

$\Rightarrow$ $r=5$

$	herefore $ Coefficient $=-{}^{9}{C}_{5}{2}^{6}=-{63.2}^{7}$

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

and $|\alpha l-\beta |=|7	imes 5+63|=98$

Solutions · 1