The term independent of x in the binomial exp | vedclass

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Question

General term of $\left(2 x^{2}-\frac{1}{x}\right)^{8}$ is

$^{8} \mathrm{C}_{\mathrm{r}}\left(2 \mathrm{x}^{2}\right)^{8-\mathrm{r}}\left(\frac{-1}{

Vedclass · Accepted Answer

$400$

Vedclass · Answer

General term of $\left(2 x^{2}-\frac{1}{x}ight)^{8}$ is

$^{8} \mathrm{C}_{\mathrm{r}}\left(2 \mathrm{x}^{2}ight)^{8-\mathrm{r}}\left(\frac{-1}{\mathrm{x}}ight)^{\mathrm{r}}$

$	herefore $ Given expression is equal to

$\left(1-\frac{1}{x}+3 x^{5}ight)^{8} C_{r}\left(2 x^{2}ight)^{8-	au}\left(-\frac{1}{x}ight)^{r}$

${ = ^8}{{	ext{C}}_{	ext{r}}}{\left( {2{{	ext{x}}^2}} ight)^{8 - {	ext{r}}}}{\left( { - \frac{1}{{	ext{x}}}} ight)^{	ext{r}}} - {\frac{1}{{	ext{x}}}^8}{{	ext{C}}_{	ext{r}}}{\left( {2{{	ext{x}}^2}} ight)^{8 - {	ext{r}}}}{\left( { - \frac{1}{{	ext{x}}}} ight)^r}$

$ + 3{{	ext{x}}^5}{ \cdot ^8}{{	ext{C}}_{	ext{r}}}{\left( {2{{	ext{x}}^2}} ight)^{8 - {	ext{r}}}}{\left( { - \frac{1}{{	ext{x}}}} ight)^{	ext{r}}}$

${ = ^8}{{	ext{C}}_{	ext{T}}}{2^{8 - {	ext{r}}}}{( - 1)^{	ext{r}}}{{	ext{x}}^{16 - 3{	ext{r}}}}{ - ^8}{{	ext{C}}_{	ext{r}}}{2^{8 - {	ext{r}}}}{( - 1)^{	ext{r}}}{{	ext{x}}^{15 - 3{	ext{r}}}}$

$ + 3{ \cdot ^8}{{	ext{C}}_{	ext{r}}}{2^{(8 - {	ext{r}})}}{\left( { - \frac{1}{{	ext{x}}}} ight)^{	ext{r}}}{( - 1)^{	ext{r}}}{{	ext{x}}^{21 - 3{	ext{r}}}}$

For the term independent of $x,$ we should have

$16-3 r=0,15-3 r=0,21-3 r=0$

From the simplification we get $r=5$ and $r=7$

$	herefore-^{8} C_{5}\left(2^{3}ight)(-1)^{5}-3 \cdot^{8} C_{7} \cdot 2$

$+\left[\frac{8 !}{5 ! 3 !} 	imes 8ight]-3 	imes\left[\frac{8 !}{7 ! 1 !} 	imes 2ight]$

$=(56	imes 8)-48$

$=448-6	imes 8=448-48=400$

The term independent of $x$ in the binomial expansion of $\left( {1 - \frac{1}{x} + 3{x^5}} \right){\left( {2{x^2} - \frac{1}{x}} \right)^8}$ is

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