If the constant term in the binomial expansio | vedclass

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Question

$	ext { In, }\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell}}ight)^9$

$T_{ r +1}={ }^9 C_{ r } \frac{\left(x^{5 / 2}ight)^{9- r }}{2^{9- r }

Vedclass · Accepted Answer

$98$

Vedclass · Answer

$	ext { In, }\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell}}ight)^9$

$T_{ r +1}={ }^9 C_{ r } \frac{\left(x^{5 / 2}ight)^{9- r }}{2^{9- r }}\left(\frac{-4}{x^{\ell}}ight)^{ r }$

$=(-1)^{ r } \frac{{ }^9 C_{ r }}{2^{9- I }} 4^{ r } x ^{\frac{45}{2}-\frac{ r }{2} Ir }$

$=45-5 r -21 r =0$

$r =\frac{45}{5+21}..........(1)$

Now, according to the question, $(-1)^{ T } \frac{{ }^9 C _{ r }}{2^{9- I }} 4^{ T }=-84$

$=(-1)^{ I }{ }^9 C _{ r } 2^{3 T -9}=21 	imes 4$

Only natural value of $r$ possible if $3 r-9=0$ $r =3$ and ${ }^9 C _3=84$

$	herefore 1=5$ from equation $(1)$

Now, coefficient of $x^{-31}=x^{\frac{45}{2}-\frac{5 r }{2}}$-ir at $l=5$, gives

$r=5$

$	herefore{ }^9 c_5(-1) \frac{4^5}{2^4}=2^\alpha 	imes \beta$

$=-63 	imes 2^7$

$\Rightarrow \alpha=7, \beta=-63$

$	herefore \quad 	ext { value of }|\alpha \ell-\beta|=98$

If the constant term in the binomial expansion of $\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell}}\right)^9$ 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

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