Binomial Theorem and its Simple Applications Previous Year Questions

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Solution

Q.
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 $x^{- 3 l}$ is $2^{α} β$ , where $β < 0$ is an odd number, then $| α l - β |$ is equal to __________ . [2023]

Ans.
(98)

$T_{r + 1} = {}^{9}C_{r} {\frac{(x^{5 / 2})}{2^{9 - r}}}^{9 - r} {(\frac{- 4}{x^{l}})}^{r}$

$= {(- 1)}^{r} \frac{{}^{9}C_{r}}{2^{9 - r}} 4^{r} x^{\frac{45}{2} - \frac{5 r}{2} - l r} \Rightarrow 45 - 5 r - 2 l r = 0$

$r = \frac{45}{5 + 2 l} \dots (i)$

$Now, according to the question, {(- 1)}^{r} \frac{{}^{9}C_{r}}{2^{9 - r}} 4^{r} = - 84$

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

$Only natural value of r possible if 3 r - 9 = 0 \Rightarrow r = 3 and {}^{9}C_{3} = 84$

$From equation (i): l = 5$

$For coefficient of x^{- 3 l} : \frac{45}{2} - \frac{5 r}{2} - l r = - 3 l \Rightarrow \frac{45}{2} - \frac{5 r}{2} - 5 r = - 3 l \Rightarrow r = 5$

$∴$ $The coefficient of x^{- 3 l} is {}^{9}C_{5} (- 1) \frac{4^{5}}{2^{4}} = 2^{α} \times β$

$\Rightarrow α = 7, β = - 63$

$∴$ $Value of$ $| α l - β |$ $= 7 \times 5 + 63 = 35 + 63 = 98$

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