In the expansion of {left( {frac{x}{2} - frac | vedclass

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Question

(a) In the expansion of ${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} \right)^{10}}$, 

the general term is ${T_{r + 1}} = {\,^{10}}{C_r}{\left( {

Vedclass · Accepted Answer

$\frac{{405}}{{256}}$

Vedclass · Answer

(a) In the expansion of ${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} ight)^{10}}$, 

the general term is ${T_{r + 1}} = {\,^{10}}{C_r}{\left( {\frac{x}{2}} ight)^{10 - r}}.\,\,{\left( { - \frac{3}{{{x^2}}}} ight)^r}$ 

${ = ^{10}}{C_r}{( - 1)^r}.\frac{{{3^r}}}{{{2^{10 - r}}}}{x^{10 - r - 2r}}$

Here, the exponent of $x$ is $10 - 3r = 4 \Rightarrow r = 2$

$	herefore \,\,\,\,{T_{2 + 1}}{ = ^{10}}{C_2}{\left( {\frac{x}{{m{2}}}} ight)^8}{\left( { - \frac{3}{{{x^2}}}} ight)^2}$

= $\frac{{10.9}}{{1.2}}.\frac{1}{{{2^8}}}{.3^2}.{x^4}$ 

= $\frac{{405}}{{256}}{x^4}$

$	herefore $ The required coefficient $ = \frac{{405}}{{256}}$.

In the expansion of ${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} \right)^{10}}$, the coefficient of ${x^4}$is

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