Mathematics - Binomial Theorem Question with Solution | TestHub

We have,${(1+2x+3x^2)}^6=\sum _{r=0}^6{ }^6{C}_r{\left(2x+3x^2\right)}^{\mathrm{r}}$
${=}^6{C}_0{+}^6{C}_1\left(2x+3x^2\right){+}^6{C}_2{\left(2x+3x^2\right)}^2+ \dots {+}^6{C}_6{\left(2x+3x^2\right)}^6$
$\therefore$Coefficient of$x^2=18+60=\mathrm{78}$
Again, coefficient of$x^2$in${\left(1-4x^2\right)}^6=-24$
Constant term in$\left({\left(1+2x+3x^2\right)}^6+{\left(1-4x^2\right)}^6\right)$is$1+1=2$.
Thus, the required coefficient of$x^2=2$(coefficient of$x^2$) in$\left({\left(1+2x+3x^2\right)}^6+{\left(1-4x^2\right)}^6\right)-$constant term in$\left({\left(1+2x+3x^2\right)}^6+{\left(1-4x^2\right)}^6\right)$
$=2\left(78-24\right)-2$
$=106$.