If the coefficients of $x^{2}$ and $x^{3}$ are both zero, in the expansion of the expression $\left(1+a x+b x^{2}\right)(1-3 x)^{15}$ in powers of $x$, then the ordered pair $(a, b)$ is equal to :
Correct Option: 1
Coefiicient of $\mathrm{x}^{2}={ }^{15} \mathrm{C}_{2} \times 9-3 \mathrm{a}\left({ }^{15} \mathrm{C}_{1}\right)+\mathrm{b}=0$
$\Rightarrow-45 \mathrm{a}+\mathrm{b}+{ }^{15} \mathrm{C}_{2} \times 9=0$ ..............(i)
Also, $-27 \times{ }^{15} \mathrm{C}_{3}+9 \mathrm{a} \times{ }^{15} \mathrm{C}_{2}-3 \mathrm{~b} \times{ }^{15} \mathrm{C}_{1}=0$
$\Rightarrow 9 \times{ }^{15} C_{2} a-45 b-27 \times{ }^{15} C_{3}=0$
$\Rightarrow 21 a-b-273=0$ .......(ii)
(i) $+$ (ii)
$-24 a+672=0$
$\Rightarrow a=28$
So, $b=315$
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