Question

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

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

Answer 1

Answer: (A)

General term of ${\left(2x^2-\frac{1}{x}\right)}^8$ is
${}^8{C}_r{\left(2x^2\right)}^{8-r}{\left(\frac{-1}{x}\right)}^r$
$\therefore$ Given expression is equal to
${\left(1-\frac{1}{x}+3x^5\right)}^8{C}_r{\left(2x^2\right)}^{8-r}{\left(-\frac{1}{x}\right)}^r$
${=}^8{C}_r{\left(2x^2\right)}^{8-r}{\left(-\frac{1}{x}\right)}^r-\frac{1}{x}{}^8{C}_r{\left(2x^2\right)}^{8-r}{\left(-\frac{1}{x}\right)}^r$
$+3x^5{.}^8{C}_r{\left(2x^2\right)}^{8-r}{\left(-\frac{1}{x}\right)}^r<br/><br/>{=}^8{C}_r{2}^{8-r}{\left(-1\right)}^r{x}^{16-3r}{-}^8{C}_r{2}^{8-r}{\left(-1\right)}^r{x}^{15-3r}$
$+3.{}^8{C}_r{2}^{\left(8-r\right)}{\left(-\frac{1}{x}\right)}^r{\left(-1\right)}^r{x}^{21-3r}$
For the term independent of $x$, we should have
$16-3r=0,15-3r=0,21-3r=0$
From the simplification we get $r=5$ and $r=7$
$\therefore {-}^8{C}_5\left(2^3\right){\left(-1\right)}^5-3.{}^8{C}_7.2+\left[\frac{8!}{5!3!}\times 8\right]-3\times \left[\frac{8!}{7!4!}\times 2\right]$
$=\left(56\times 8\right)-48$
$=448-6\times 8=448-48=400$