Mathematics - Binomial Theorem Question with Solution | TestHub

$\left(1+ax+b{x}^2\right){\left(1-3x\right)}^{15}=\left(1+ax+b{x}^2\right)\left({}^{15}C_0{-}^{15}{C}_1 3x {+}^{ 15}{C}_2 {\left(3x\right)}^2{-}^{15}{C}_3{\left(3x\right)}^3+\dots {-}^{15}{C}_{15}{\left(3x\right)}^{15}\right)$

$\therefore$ Coefficient of $x^2$ is: ${ b.}^{15}{C}_0-3.a^{15}{C}_1+3^{2.}{.}^{15}{C}_2=0$

(Given)
$\Rightarrow b-45a+945=0......\left(\mathrm{i}\right)$

Coefficient of $x^3$ is:

$-b.3{.}^{15}{C}_1+9a{.}^{15}{C}_2-27{.}^{15}{C}_3=0$

$\Rightarrow -45b+9a\times \frac{15}{2}\times \frac{14}{1}-27\times \frac{15}{3}\times \frac{14}{2}\times \frac{13}{1}=0$

$\Rightarrow -b+21a-273=0$

$\Rightarrow b-21a+273=0 .......\left(\mathrm{ii}\right)$

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$

$a=28$ and $b=315$