The coefficient of x^2 in the expansion | vedclass

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Question

$	ext { Let } a=\left(\left(1+2 x+3 x^{2}ight)^{6}+\left(1-4 x^{2}ight)^{6}ight)$

$	herefore $ Coefficient of $x^{2}$ in the expansion of t

Vedclass · Accepted Answer

$106$

Vedclass · Answer

$	ext { Let } a=\left(\left(1+2 x+3 x^{2}ight)^{6}+\left(1-4 x^{2}ight)^{6}ight)$

$	herefore $ Coefficient of $x^{2}$ in the expansion of the product

$\left(2-x^{2}ight)\left(\left(1+2 x+3 x^{2}ight)^{6}+\left(1-4 x^{2}ight)^{6}ight)$

$=2\left(	ext { Coefficient of } x^{2} 	ext { in a }ight)-1$ (Constant of expansion)

In the expansion of

$\left(\left(1+2 x+3 x^{2}ight)^{6}+\left(1-4 x^{2}ight)^{6}ight)$

Constant $=1+1=2$

Coefficient of $x^{2}=$

[Coefficient of $x^2$ in ${(^6}{C_0}{(1 + 2x)^6}{(3{x^2})^0}$ ] $+$ [Coefficient of $x^2$ in ${(^6}{C_1}{(1 + 2x)^5}{(3{x^2})^1}$ ] $-$ ${[^6}{C_1}(4{x^2})]$ 

$=60+6 	imes 3-24=54$

$	herefore \quad$ The coefficient of $x^{2}$ in $\left(2-x^{2}ight)$

${\left(\left(1+2 x+3 x^{2}ight)^{6}+\left(1-4 x^{2}ight)^{6}ight)}$

${=2 	imes 54-1(2)=108-2=106}$

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

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