The coefficient of x^4 in left[ frac{x}{2} - | vedclass

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(A) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.For the expression $\left[ \frac{x}{2}

Vedclass · Accepted Answer

$\frac{405}{256}$

Vedclass · Answer

(A) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.For the expression $\left[ \frac{x}{2} - \frac{3}{x^2} ight]^{10}$,we have $a = \frac{x}{2}$,$b = -\frac{3}{x^2}$,and $n = 10$.$T_{r+1} = {}^{10}C_{r} \left( \frac{x}{2} ight)^{10-r} \left( -\frac{3}{x^2} ight)^{r}$$T_{r+1} = {}^{10}C_{r} \cdot \frac{x^{10-r}}{2^{10-r}} \cdot (-1)^r \cdot \frac{3^r}{x^{2r}}$$T_{r+1} = {}^{10}C_{r} \cdot (-1)^r \cdot \frac{3^r}{2^{10-r}} \cdot x^{10-3r}$To find the coefficient of $x^4$,we set the exponent of $x$ equal to $4$:$10 - 3r = 4$$3r = 6$$r = 2$Substituting $r=2$ into the expression for the coefficient:Coefficient $= {}^{10}C_{2} \cdot (-1)^2 \cdot \frac{3^2}{2^{10-2}}$$= \frac{10 	imes 9}{2 	imes 1} \cdot 1 \cdot \frac{9}{2^8}$$= 45 \cdot \frac{9}{256} = \frac{405}{256}$

The coefficient of $x^4$ in $\left[ \frac{x}{2} - \frac{3}{x^2} \right]^{10}$ is:

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