If the constant term in the binomial expansio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The general term $T_{r+1}$ in the expansion of $\left(\frac{x^{5/2}}{2} - \frac{4}{x^{\ell}}\right)^9$ is given by:$T_{r+1} = \binom{9}{r} \left(\

Vedclass · Accepted Answer

$98$

Vedclass · Answer

(D) The general term $T_{r+1}$ in the expansion of $\left(\frac{x^{5/2}}{2} - \frac{4}{x^{\ell}}ight)^9$ is given by:$T_{r+1} = \binom{9}{r} \left(\frac{x^{5/2}}{2}ight)^{9-r} \left(-\frac{4}{x^{\ell}}ight)^r$$T_{r+1} = \binom{9}{r} \frac{1}{2^{9-r}} (-4)^r x^{\frac{5(9-r)}{2} - r\ell}$For the constant term,the exponent of $x$ must be $0$:$\frac{45-5r}{2} - r\ell = 0 \implies 45 - 5r - 2r\ell = 0 \implies r(5 + 2\ell) = 45$Given the constant term is $-84$:$(-1)^r \binom{9}{r} \frac{4^r}{2^{9-r}} = -84$$(-1)^r \binom{9}{r} 2^{2r - 9 + r} = -84 \implies (-1)^r \binom{9}{r} 2^{3r-9} = -84$For $r=3$,$\binom{9}{3} = 84$. Thus,$(-1)^3 (84) 2^{3(3)-9} = -84 	imes 2^0 = -84$. This matches.Substituting $r=3$ into $r(5+2\ell) = 45$:$3(5+2\ell) = 45 \implies 5+2\ell = 15 \implies 2\ell = 10 \implies \ell = 5$.Now,find the coefficient of $x^{-3\ell} = x^{-15}$:$\frac{45-5r}{2} - 5r = -15 \implies 45 - 5r - 10r = -30 \implies 15r = 75 \implies r = 5$.The coefficient is $(-1)^5 \binom{9}{5} \frac{4^5}{2^{9-5}} = -126 	imes \frac{2^{10}}{2^4} = -126 	imes 2^6 = -63 	imes 2^7$.Given $2^{\alpha}\beta = -63 	imes 2^7$,we have $\alpha = 7$ and $\beta = -63$.Finally,$|\alpha\ell - \beta| = |7(5) - (-63)| = |35 + 63| = 98$.

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

Similar Questions

If the second,third,and fourth terms in the expansion of $(x + a)^n$ are $240, 720,$ and $1080$ respectively,then the value of $n$ is

Let $\alpha$ be the constant term in the binomial expansion of $\left(\sqrt{x} - \frac{6}{x^{3/2}}\right)^n$,$n \leq 15$. If the sum of the coefficients of the remaining terms in the expansion is $649$ and the coefficient of $x^{-n}$ is $\lambda \alpha$,then $\lambda$ is equal to $..........$.

The number of rational terms in the binomial expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is

If the $1011^{\text{th}}$ term from the end in the binomial expansion of $(\frac{4x}{5} - \frac{5}{2x})^{2022}$ is $1024$ times the $1011^{\text{th}}$ term from the beginning,then $|x|$ is equal to

If the coefficient of $x^7$ in $(ax - \frac{1}{bx^2})^{13}$ and the coefficient of $x^{-5}$ in $(ax + \frac{1}{bx^2})^{13}$ are equal,then $a^4 b^4$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module