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

34

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107

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155

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108

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Solution

The correct option is A $34$
$2 (1 + 2 x + 3 x^{2})^{6} - x^{2} (1 + 2 x + 3 x^{2})^{6} + 2 (1 - 4 x^{2})^{6} - x^{2} (1 - 4 x^{2})^{6}$
$↓$ $↓$ $↓$ $↓$
Coeff of $x^{2}$ Coeff of $x^{0}$ Coeff of $x^{2}$ Coeff of $x^{0}$

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

$\propto_{2} + 2 \propto_{3} = 2$ $\propto_{2} + 2 \propto_{3} = 0$

$\propto_{3} = 1, \propto_{2} = 2, \propto_{1} = 4$

$\propto_{2} = 0, \propto_{3} = 0, \propto_{1} = 6$

$\propto_{3} = 1, \propto_{2} = 0, \propto_{1} = 5$

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

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

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

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

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

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

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