Question

Find the coefficient of $x^2$ in the expansion of $(1+x^2){(\frac{x}{2}-\frac{4}{x})}^6$

Answer 1

$(1+x^2){(\frac{x}{2}-\frac{4}{x})}^6=T_1.T_2$

Binomial expansion of $T_2={(\frac{x}{2}-\frac{4}{x})}^6$

${\sum }_{r=0}^{r=6}{}^6{C}_r{(-1)}^r{\left(\frac{x}{2}\right)}^r{\left(\frac{4}{x}\right)}^{6-r}$

$={\sum }_{r=0}^{r=6}{}^6{C}_r{(-1)}^r{2}^{-r}{4}^{6-r}{x}^{r-6+r}$

$={\sum }_{r=0}^{r=6}{}^6{C}_r{(-1)}^r{2}^{-r}{2}^{12-2r}{x}^{2r-6}$

$={\sum }_{r=0}^{r=6}{}^6{C}_r{(-1)}^r{2}^{12-3r}{x}^{2r-6}$

To find coefficient of $x^2$

This can be done in $2$ ways:

$1.$Constant term of $T_1$ is multiplied by the coefficient of $x^2$ of $T_2$

$\Rightarrow 2r-6=2\Rightarrow r=4$

Constant of $T_1=1$

Coefficient of $x^2$ is $T_2$

Substituting $r=4$ in ${}^6{C}_r{(-1)}^r{2}^{12-3r}$

${}^6{C}_4{(-1)}^4{2}^{12-3\times 4}=\frac{6\times 5}{1\times 2}\times 1\times 2^0=15$

Coefficient of $x^2=$constant of $T_1\times$Coeffcient of $x^2$ of $T_2$

$=C_1=1\times 15=15$

$x_2$ of $T_1$ is multiplied by constant term of $T_2$

Coefficient of $x_2$ of $T_1=1$

Constant term of $T_2=$ when $x$ is raised to the power of $0$

$\Rightarrow 2r-6=0\Rightarrow r=3$

Substituting $r=3$ in ${}^6{C}_r{(-1)}^r{2}^{12-3r}$

$={}^6{C}_3{(-1)}^3{2}^{12-3\times 3}$

$=\frac{6\times 5\times 4}{1\times 2\times 3}(-1)2^3=-160$

Coefficient of $x^2=$ Coefficient of $x^2$ of $T_1\times$ Constant of $T_2=C_2=15-160=-145$