# The term independent of x in the binomial expansion ofis: Option 1) 400 Option 2) 496 Option 3) -400 Option 4) -496

As we learnt in

Independent term -

A term which does not contains x, then

$\dpi{120} x^{\left ( n,r \right )}= x^{0}$

$\dpi{120} \therefore \left ( n,r \right )= 0$ find $\dpi{120} r$

- wherein

Where $\dpi{120} r\geqslant 0$

$\left(1-\frac{1}{x}+3x^{5} \right )\left(2x^{2}-\frac{1}{x} \right )^{8}$

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

$\Rightarrow\ \; ^{8}C_{r}\ 2^{8-r},\ x^{16-2r-r}\ x(-1)^{r}$

$\Rightarrow\ \; ^{8}C_{r}\ 2^{8-r},\ x^{16-3r}\ x(-1)^{r}$

We need the powers x0, x1, x-5

$16-3r=0\ \;\Rightarrow\ \; r=\frac{16}{3},$ inot possible.

$16-3r=1\ \;\Rightarrow\ \; r=5$,

For n = 5, we get $^{8}C_{5}\ 2^{3} (-1)^{5}=\frac{-8\times 8\times 7\times 6}{6}=-448$

If 16 - 3r = -5

r = 7, we get $^{8}C_{7}\ 2^{1}. (-1)^{1}=-16$

SQ we get  $\left(1-\frac{1}{x}+3x^{5} \right )\left(-448x-16^{-5} \right )=448-48=400$

Correct option is 1.

Option 1)

400

This is the correct option.

Option 2)

496

This is an incorrect option.

Option 3)

-400

This is an incorrect option.

Option 4)

-496

This is an incorrect option.

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