Show that the coefficient of x4 in the expansion of

Question:

Show that the coefficient of $x^{4}$ in the expansion of $\left(\frac{x}{2}-\frac{3}{x^{2}}\right)^{10}$ is $\frac{405}{256}$

 

Solution:

To show: that the coefficient of $x^{4}$ in the expansion of $\left(\frac{x}{2}-\frac{3}{x^{2}}\right)^{10}$ is -330.

Formula Used:

General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where

${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{r !(n-r) !}$

Now, finding the general term of the expression $\left(\frac{\mathrm{x}}{2}-\frac{3}{\mathrm{x}^{2}}\right)^{10}$ we get

$T_{r+1}={ }^{10} C_{r} \times\left(\frac{x}{2}\right)^{10-r} \times\left(\frac{-3}{x^{2}}\right)^{r}$

For finding the term which has ${x}^{4}$ in it, is given by

$10-3 r=4$

$3 r=6$

$R=2$

Thus, the term which has $\mathrm{x}^{4}$ in it is $\mathrm{T}_{3}$

$T_{3}={ }^{10} C_{2} \times\left(\frac{x}{2}\right)^{8} \times\left(\frac{-3}{x^{2}}\right)^{2}$

$T_{3}=\frac{10 ! \times 9}{2 ! \times 8 ! \times 2^{8}}$

$T_{3}=\frac{10 \times 9 \times 8 ! \times 9}{2 \times 8 ! \times 2^{8}}$

$T_{3}=\frac{405}{256}$

Thus, the coefficient of $x^{4}$ in the expansion of $\left(\frac{x}{2}-\frac{3}{x^{2}}\right)^{10}$ is $\frac{405}{256}$

 

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