Question

The coefficient of $x^2$ in the expansion of the product $(2-x^2).({(1+2x+3x^2)}^6+{(1-4x^2)}^6)$ is

• A. $34$
• B. $107$
• C. $155$
• D. $108$

Answer 1

The correct option is A $34$

$2(1+2x+3x^2{)}^6-x^2(1+2x+3x^2{)}^6+2(1-4x^2{)}^6-x^2(1-4x^2{)}^6$

$\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$

Coeff of $x^2$ Coeff of $x^0$ Coeff of $x^2$ Coeff of $x^0$

$(1+2x+3x^2{)}^6\Rightarrow \frac{6!}{{\propto }_1!{\propto }_2!{\propto }_3!}(2n{)}^{\propto 2}(3x^2{)}^{\propto 3}\Rightarrow \frac{6!}{{\propto }_1!{\propto }_2!{\propto }_3!}{2}^{\propto 2}{3}^{\propto 3}{x}^{\propto 2+2\propto 3}$

${\propto }_2+2{\propto }_3=2$ ${\propto }_2+2{\propto }_3=0$

${\propto }_3=1,{\propto }_2=2,{\propto }_1=4$

${\propto }_2=0,{\propto }_3=0,{\propto }_1=6$

${\propto }_3=1,{\propto }_2=0,{\propto }_1=5$

$2(\frac{6!}{2!4!}{2}^2+\frac{6!}{1!5!}{2}^0{3}^1)-1\left(\frac{6!}{6!}{2}^0{3}^0\right)$

$(1-4x^2{)}^6{=}^6{C}_r(-4x^2{)}^r$

${=}^6{C}_r(-4{)}^r{x}^{2r}$

for coeff. of $x^{2}r=1$

${\Rightarrow }^6{C}_1(-4{)}^1$

for ecoeff of $x^0,r=0$

${\Rightarrow }^6{C}_0(-4{)}^0$

coeff of $x^2\Rightarrow \frac{6!}{2!4!}\times 8+\frac{6!}{1!5!}\times 6-\frac{6!}{6!}+2\times \frac{6!}{5!}(-4)-\frac{6!}{6!}$

$\Rightarrow \frac{6\times 5\times 8}{2}+6\times 6-1-2\times 4\times 6-1$

$\Rightarrow 12-36-1-48-1\Rightarrow 34$