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
$\left(1-\frac{1}{x}+3x^{5}\right)^{8}C_{r}\left(2x^{2}\right)^{8-r}\left(-\frac{1}{x}\right)^{r}$
$=^{8}C_{r}\left(2x^{2}\right)^{8-r}\left(-\frac{1}{x}\right)^{r}-\frac{1}{x} \,{}^{8}C_{r}\left(2x^{2}\right)^{8-r}\left(-\frac{1}{x}\right)^{r}$
$+3x^{5}. ^{8}C_{r}\left(2x^{2}\right)^{8-r}\left(-\frac{1}{x}\right)^{r}
=^{8}C_{r}2^{8-r}\left(-1\right)^{r}\,x^{16-3r}-^{8}C_{r}\,2^{8-r}\left(-1\right)^{r}\,x^{15-3r}$
$+3.\,{}^{8}C_{r}2^{\left(8-r\right)}\left(-\frac{1}{x}\right)^{r}\,\left(-1\right)^{r}\,x^{21-3r}$
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 $r=7$
$\therefore -^{8}C_{5}\left(2^{3}\right)\left(-1\right)^{5}-3.\,{}^{8}C_{7}.2+\left[\frac{8!}{5!3!}\times8\right]-3\times\left[\frac{8!}{7!4!}\times2\right]$
$=\left(56\times8\right)-48$
$=448-6\times8 = 448-48 = 400$