Question

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

• A. $-496$
• B. $-400$
• C. $496$
• D. $400$

Answer 1

The correct option is D $400$

Given expansion of $(1-\frac{1}{x}+3x^5){(2x^2-\frac{1}{x})}^8$

For this, we write general term in the expansion of ${(2x^2-\frac{1}{x})}^8$

$T_{r+1}{=}^8{C}_r\left(2x^2{)}^{8-r}{\left(\frac{1}{x}\right)}^{-r}{=}^8{C}_r{2}^{8-r}\right(x^{16-3r})$

Now, in the product $(1-\frac{1}{x}+3x^5){(2x^2+(-\frac{1}{x})}^8$, the term independent of $x$ is

$1\times$ term not containing $x$ in ${(2x^2-\frac{1}{x})}^8$ $-\frac{1}{x}\times$ term containing $x$ in ${(2x^2-\frac{1}{x})}^8$ $+3x^5\times$ term containing $x^{-5}$ in

${(2x^2-\frac{1}{x})}^8$ .....(2)

Now, by (1), term without $x$ ,

$16-3r=0$

$\Rightarrow r=\frac{16}{3}$ which is not possible.

Hence, there is no term independent of $x$ in ${(2x^2-\frac{1}{x})}^8$

Now, again by (1), term containing $x$

$16-3r=1$

$\Rightarrow r=5$

So, by (1), $T_6=-448x$

Now, again by (1), term containing $x^{-5}$

$16-3r=-5$

$\Rightarrow r=7$

So, by (1), $T_8=-16x^{-5}$

Put all these values in (2), we get

Term independent of $x=1\times 0$ $-\frac{1}{x}(-448x)$ $+3x^5(-16x^{-5})$

$=448-48=400$